Download Quantum Mechanics and Number Theory

Quantum Mechanics and Number Theory: an
Introduction
(preliminary version)
Giuseppe Mussardo
International School for Advanced Studies (SISSA)
via Beirut 2-4, 34014 Trieste, Italy
and
Piergiulio Tempesta
Departamento de Fı́sica Teórica II,
Métodos y modelos matemáticos,
Facultad de Fı́sicas, Universidad Complutense,
28040 – Madrid, Spain
June 3th, 2010
2
PREFACE
These Lecture notes are a first version of a book in progress on Quantum
Mechanics and Number Theory. This is a living book: the present version
is still preliminary and will be periodically updated.
The figures will be provided in a separate file.
Contents
I
Quantum mechanics
7
1 Mathematical Introduction
1.1 Formulas for π . . . . . . . . . . . .
1.1.1 Leibnitz formula . . . . . . .
1.1.2 Viete’s formula . . . . . . . .
1.1.3 Other formulas for π . . . . .
1.2 The quantum propagator . . . . . .
1.3 spectrum depending on one quantum
1.4 The Gamma Function . . . . . . . .
1.5 Poisson summation formula . . . . .
1.6 Spectral Functions . . . . . . . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
number
. . . . .
. . . . .
. . . . .
2 WKB Approximation
2.1 Introduction . . . . . . . . . . . . . . . . . .
2.2 The connection formulas . . . . . . . . . . .
2.3 Infinite well . . . . . . . . . . . . . . . . . .
2.4 Airy function . . . . . . . . . . . . . . . . .
2.5 WKB for radial motion . . . . . . . . . . .
2.6 Square–well potential . . . . . . . . . . . . .
2.7 connection formulas . . . . . . . . . . . . .
2.8 Transmission through a barrier . . . . . . .
2.9 Metastable states . . . . . . . . . . . . . . .
2.10 Potential V (x) = λ |x|a . . . . . . . . . . .
2.10.1 WKB quantization of V (x) = λ |x|a
2.11 semiclassical estimate of hxα in . . . . . . .
2.12 Exact WKB quantization . . . . . . . . . .
2.12.1 Some ”numerology” . . . . . . . . .
2.13 Estimate of the ground state energy . . . .
2.14 Uncertainty relation and lower bound . . .
3
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9
9
9
12
13
15
22
24
29
33
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37
37
39
42
42
44
44
47
50
50
51
52
53
56
59
60
62
4
CONTENTS
3 TBA and differential equations
65
3.1 Spectral determinant . . . . . . . . . . . . . . . . . . . . . . . 65
3.2 Hadamard Factorization Theorem . . . . . . . . . . . . . . . 72
3.3 Bethe Ansatz Equations . . . . . . . . . . . . . . . . . . . . . 74
II
Analytic Number Theory: an Introduction
77
4 Prime numbers
79
4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79
4.2
4.3
4.4
4.5
4.6
4.7
4.8
4.9
4.10
4.11
Induction . . . . . . . . . . . . . . . . . . . .
Prime Numbers . . . . . . . . . . . . . . . . .
Measure of Complexity . . . . . . . . . . . . .
A Primality Test in Quantum Mechanics . . .
Factorization Problem in Quantum Mechanics
Elementary facts about prime numbers . . . .
Fermat’s little theorem . . . . . . . . . . . . .
Prime numbers and polynomials . . . . . . .
Distribution of prime numbers . . . . . . . .
Probabilistic methods . . . . . . . . . . . . .
4.11.1 Coprime probability . . . . . . . . . .
4.11.2 Square–free probability . . . . . . . .
4.11.3 Merten’s formula . . . . . . . . . . . .
4.12 Mersenne numbers . . . . . . . . . . . . . . .
4.13 Arithmetical functions . . . . . . . . . . . . .
4.14 Dirichlet characters . . . . . . . . . . . . . . .
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81
82
83
84
84
86
86
88
89
92
92
92
93
93
95
95
5 The
5.1
5.2
5.3
5.4
Riemann ζ function
Q
Relation between ζ (s) and (x) . . . . . . . .
Moebius function . . . . . . . . . . . . . . . . .
The Mangoldt function Λ (n) . . . . . . . . . .
Analyticity properties of ζ (s) . . . . . . . . . .
5.4.1 Consequences of the functional equation
Some consequences of Euler’s formula . . . . .
Dirichlet series . . . . . . . . . . . . . . . . . .
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97
98
99
100
101
104
105
107
5.5
5.6
6 Numerical Spectral Methods
109
6.1 Basic formulas for the quantum harmonic oscillator . . . . . . 109
6.2 Truncated Hilbert space method . . . . . . . . . . . . . . . . 111
6.3 Tight–Binding Method . . . . . . . . . . . . . . . . . . . . . . 114
CONTENTS
5
6.3.1
Particle in a box . . . . . . . . . . . . . . . . . . . . . 114
6
CONTENTS
Part I
Quantum mechanics
7
Chapter 1
Mathematical Introduction
In this first Lecture, we will revise some basic aspects of Elementary Number
Theory.
1.1
Formulas for π
1.1.1
Leibnitz formula
The following result, due to Leibnitz, holds:
∞
X (−1)n
π
1 1 1
= 1 − + − + ... =
.
4
3 5 7
2n + 1
n=0
The proof is immediate, since
π
= arctg1 =
4
Z
0
1
∞
X
dx
=
1 + x2
Z
1
n=0 0
∞
X
(−1)n
(−1) x dx =
.
2n + 1
n
2n
n=0
The above formula is completely equivalent to the one of Lord Bronncker
4
12
=1+
32
π
2+
52
2+
.
2
2+ 7
..
.
This remarkable formula links π with the integers in a far more striking way
that does the decimal expansion of π, which appears to exhibit no regularity
in the sequence of its digits.
9
10
CHAPTER 1. MATHEMATICAL INTRODUCTION
The equivalence between the two formulas is based on the fact that a convergent infinite series of the form
γ1 + γ1 γ2 + γ1 γ2 γ3 + γ1 γ2 γ3 γ4 + ...
is equivalent to the continued fraction
γ1
1−
γ2
γ
1+γ2 − 1+γ3 −
.
3
Other remarkable continued fractions.
1+
1+
1
1+
1
1+
s
1
1+
e−2π
−4π
1+ e
√
5+1
=
.
2
1
=
..
.

√
√
5+ 5
5 + 1  2π
−
e5.
2
2
..
.
The Leibnitz formula is however slowly convergent. Better formulas are, for
instance
∞
∞
π2 X 1
π4 X 1
=
=
,
.
6
n2
90
n4
1+
n=1
n=1
As we will see, we also have the general formula
ξ (2k) =
∞
X
1
(2π)2k
(−1)k−1 B2k ,
=
n2k
2 (2k!)
n=1
where B2k are the (even) Bernoulli numbers, defined as the coefficients of
the expansion
∞
X
x
xn
=
B
.
n
ex − 1
n!
n=0
They satisfy the recursive equation
Bn+1 =
n+1
Xµ
k=0
n+1
k
¶
Bk ,
which is one of the ways for calculating them. Their first values are
B0 = 1,
1
B1 = − ,
2
1
B2 = ,
6
B4 = −
1
...
30
1.1. FORMULAS FOR π
11
with B2k+1 = 0 for k = 1, 2, 3, ...
Exercise
Prove that
∞
7π 4 X (−1)n
=
.
720
n4
n=1
Another relevant formula, due to Wallis, is
π
2 4 4 6 6 8 8
= · · · · · · · ...
4
3 3 5 5 7 7 9
It can be proved in different ways.
1) The first proof is based on the infinite product expansion of sin x
∞ ·
³ x ´2 ¸
Y
.
sin x = x
1−
kπ
k=1
Substituting x = π2 , we have
2
1·33·55·7
=
...
π
2·24·46·6
2) Another proof is based on the integral
Z
1¡
1 − x2
I (n) =
¢n
dx.
0
It is easy to see that it satisfies the recursive equation
I (n) =
2n
I (n − 1) .
2n + 1
Hence, if n is an integer, we have
2
2
I (1) = I (0) = ,
3
3
4
2·4
I (2) = I (1) =
,
5
3·5
...
2 · 4 · 6 · ... · 2n
I (n) =
.
3 · 5 · 7 · ... · (2n + 1)
Consider now the sequence of I (n) for half integers
1 3 5
n = , , , ...
2 2 2
12
CHAPTER 1. MATHEMATICAL INTRODUCTION
where
µ ¶ Z 1p
1
π
I
=
1 − x2 dx = ,
2
4
0
...
¶
µ
1
1 · 3 · 5... (2n − 1) π
I n−
=
.
2
2 · 4 · 6...2n
2
But it is easy to prove that
µ
¶
1
I (n) ≤ I n −
≤ I (n + 1) ,
2
i.e.
2 · 4 · 6 · ... · 2n
1 · 3 · 5... (2n − 1) π
2 · 4 · 6 · ... · (2n − 2)
≤
≤
,
3 · 5 · 7 · ... · (2n + 1)
2 · 4 · 6...2n
2
3 · 5 · 7 · ... · (2n + 1)
so that one obtains Wallis’ result for n → ∞.
1.1.2
Viete’s formula
Another beautiful formula is
q
p
√
√ p
√
2
2 2+ 2 2+ 2+ 2
=
...
π
2
2
2
This formula was found by Viete (2540–160..) at the end of the 16th century.
It is the first numerical formula for π and the first one expressed as an infinite
product. The proof is based on the duplication formulas of trigonometric
functions:
sin θ = 2 sin
θ
θ
θ
θ
θ
θ
θ
θ
θ
cos = 4 sin cos cos = ... = 2n sin n cos cos ... cos n .
2
2
4
4
2
2
2
4
2
Since
θ
= θ,
n→∞
2n
q
1
taking θ = π2 and using the formula cos x2 =
2 (1 + cos x), we arrive to
Viete’s formula.
Another proof is based on the recursive formula
q
p
a2n = 2 − 4 − a2n ,
lim 2n sin
which relates the length an of the side of a regular n–gon inscribed in the
unit circle to the side a2n of the 2n–gon.
1.1. FORMULAS FOR π
1.1.3
13
Other formulas for π
In the formula
π=
∞ ·
X
n=0
4
2
1
1
−
−
−
8n + 1 8n + 4 8n + 5 8n + 6
¸µ
1
16
¶n
1
the factor 16
appears. Since this turns out to be the anomalous dimension of
the spin–field of the Ising Model, on would speculate on the field theoretical
origin of this number.
The following surprising formula is due to Ramanujan:
√ ∞
1
8 X (4n)! [1103 + 26390n]
.
=
π
9801
(n!)4 (396)4n
n=0
The fastest convergent formula is the one found by Gauss, and related to the
arithmetic–geometric mean (AGM) of two numbers (a, b). This is defined
as the common limit of the sequences
½
an+1 = √
(an+bn ) /2
a0 = a
.
bn+1 = an bn
b0 = b
It is easy to see that {an } decreases whereas {bn } increases, as it comes from
the inequalities
p ´2
1 ³√
an+1 − bn+1 =
an − bn ≥ 0,
2
1
(an − bn ) ≥ 0,
2
p ³√
p ´
bn+1 − bn = bn
an − bn ≥ 0.
an − an+1 =
Hence these sequences converge. From the first equality, we have in fact
an − bn ≤ 2−n ,
i.e. the limit is reached exponentially fast. The AGM gives rise to the
highly efficient method to compute π. Jonathan and Peter Borwein have
found surprising formulas, like
Ã
!
h
³√ ´i2
X
n
π = 2 AGM
2, 1
/ 1−
2 εn , εn = a2n − b2n .
n
14
CHAPTER 1. MATHEMATICAL INTRODUCTION
There is also a formula which links π to the Fibonacci numbers, defined by
Fn+2 = Fn+1 + Fn ,
F1 = F2 = 1.
There is a close expression for these numbers given by the Binet formula:
"Ã √
!n Ã
√ !n #
1
5+1
1− 5
Fn = √
−
,
2
2
5
simple to prove. In fact, from the ansatz Fn = xn , the recursive equation
gives
√
1± 5
2
x =x+1⇒x=
,
2
hence
Fn = Axn+ + Bxn− ,
where the two constants A and B can be fixed by the initial conditions:
A = −B = √15 .
The Fibonacci numbers are generated by
µ
¶n µ
¶
1 1
Fn+1 Fn
=
,
1 0
Fn
Fn−1
from which we infer
Fn−1 Fn+1 − Fn2 = (−1)n .
Notice that the Fibonacci numbers admit a generating function, given by
∞
X
x
=
Fn xn
1 − x − x2
n−1
(Generating function of a partition
∞
Y
n=1
q
with P (n) '
1√ π
e
4n 3
2n
3
∞
X
1
=
P (n) q n ,
1 − qn
n=0
. Finally, the formula connecting Fn with π is
∞
1
π X
=
arctan
.
4
F2n+1
n=1
1.2. THE QUANTUM PROPAGATOR
1.2
15
The quantum propagator
A basic quantity in Quantum Mechanics is given by the quantum propagator.
This is defined as the amplitude for a particle that is in a position q0 at the
time t0 to reach the point q at the time t:
K (q, t; q0 , t0 ) = hqt| q0 t0 i .
In the following, we will treat for simplicity the one dimensional case, but
the relevant formulas can be easily generalized to higher dimensions (as a
matter of fact, they can be generalized to Quantum Field Theory as well).
We will assume that at any given time, the coordinate operator qb possesses
a complete set of eigenvalues, so that
qb (t) |q, ti = q |q, ti ,
with the completeness condition
Z
dq |q, ti hq, t| = 1.
There are two ways of expressing the quantum propagator. The first one
employs wave functions, the second one the concept of “path integral”.
First of all, the translation in time is dictated by the Hamiltonian, so that
i
K (q, t; q0 , t0 ) = hq| e− ~ H(t−t0 ) |q0 i .
Inserting now a complete set {|En i}of orthogonal eigenstates of H, we have
X
i
K (q, t; q0 , t0 ) =
hq| En i hEn | e− ~ H(t−t0 ) |Em i hEn | q0 i =
n,m
X
i
ψn (q) ψn∗ (q0 ) e− ~ En (t−t0 ) ,
n
where
hq| En i = ψn (q) , hEn | qi = ψn∗ (q) .
where we have used the orthogonality of the eigenstates. The above is the
first expression of the quantum propagator. To get the other one, let’s
divide initially the time interval T = (t − t0 ) in n small intervals (n → ∞),
and let’s insert (n − 1) times the completeness relation:
Z
hqt| q0 t0 i = dq1 ...dqn hqt| qn−1 tn−1 i hqn−1 tn−1 | qn−2 tn−2 i ... hq1 t1 | q0 t0 i .
(1.1)
16
CHAPTER 1. MATHEMATICAL INTRODUCTION
Let us consider each term in this expression.
iδt
hqk tk | qk−1 tk−1 i = hqk | e− ~ H |qk−1 i =
Z
iδt
dpk−1 hqk | pk−1 i hpk−1 | e− ~ H |qk−1 i ,
(1.2)
where we have introduced a completeness relation in term of the momentum
eigenstates:
Z
dp |pi hp| = 1.
Suppose that the Hamiltonian is given by
H=
p2
+ V (q) .
2m
Using the Baker–Campbell–Hausdorff formula at the leading order in δt, we
have
iδt
iδt p2
iδt
hpk−1 | e− ~ H |qk−1 i ' hpk−1 | e− ~ 2m e− ~ V (q) |qk−1 i =
·
hpk−1 | qk−1 i e
Since
− iδt
~
p2
+V
2m
¸
(qk−1 )
.
1
hq| pi = √
eipq/~ .
2π~
For the expression (1.2) we have
1
hqk tk | qk−1 tk−1 i '
2π~
Z
dpk−1 e−
2
i(q −q
)pk−1
iδt pk−1
+ k k−1
~ 2m
~
e−
iδt
~
·
.
In the infinitesimal time δt, we may assume a constant velocity q k−1 , such
·
that qk − qk−1 ' q k−1 δt, and we can perform the Gaussian integral. The
result is very simple:
r
·
¸
m
δt ³ · ´
hqk tk | qk−1 tk−1 i =
exp
L q, q ,
2πi~δt
~
³ ·´
where L q, q is the Lagrangian of the theory:
·2
mq
− V (q) .
L=
2
1.2. THE QUANTUM PROPAGATOR
17
Putting now this result into (1.1), we arrive to the interesting result for the
quantum propagator:
³ ·´
Z
R
i/~ tt L q,q dτ
0
k (q, t; q0 , t0 ) = Dqe
,
(1.3)
where the measure is formally defined by
³ mn ´n/2
dq1 ...dqn−1 .
n→∞ 2πi~T
Dq = lim
The meaning of the above expression is obvious.. One can go from the
point (q0 , t0 ) to the point (q, t) by following all possible paths. These paths,
however, are weighted with the oscillating phase of the action of these paths.
In the semiclassical limit ~ → 0, the above integral can be evaluated in the
saddle point approximation, i.e. there exists one path which dominates the
expression. This is given by the condition
Z t ³ ´
·
δ
L q, q dτ = 0,
t0
i.e.
Z t"
t0
#
#
Z t"
∂L
∂L
∂L ·
d ∂L
t
δq + · δ q dt =
−
· δqdt + pδq |t0 .
∂q
∂q
dt
t0
∂q
∂q
(1.4)
For variational paths which leave the end points fixed, the last term in (17)
vanishes. Therefore we see that in the semiclassical limit the Lagrangian
satisfies the classical equation of motion
∂L
d ∂L
−
= 0.
∂q
dt ∂ q·
It is now immediate to see that K (q, t, q0 , t0 ) is the Green function of the
time–dependent Schrödinger operator, i.e. the propagator of the theory. In
fact, from the time evolution of the states
i
|ψ (t)i = e− ~ H(t−t0 ) |ψ (t0 )i
we can “sandwich” it with hq|, obtaining
Z
Z
¯ 0E D 0¯
³
´ ³ 0 ´
i
0
0
0
0
¯
¯
hq| ψ (t)i = ψ (q, t) = dq hqk | e− ~ H(t−t0 ) ¯q
q ¯ ψi = dq K q, t, q0 , t0 ψ q , t0 .
18
CHAPTER 1. MATHEMATICAL INTRODUCTION
The function K (q, q0 , t − t0 ) satisfies the composition law
³ 0
´ Z
³ 00
´ ³ 00 0 00
´
00
00
0
0
q, q , t − t = dq K q, q , t − t K q , q , t − t ,
and the differential equation
µ
¶ ³
´
³
´
∂
0
0
0
−i~ + H K q, q , t − t = −i~δ q − q .
∂t
We want to consider now the density of states of the system. This is defined
as
X
g (E) =
δ (E − En ) gn ,
n
where gn is the degeneracy of the energy levels. We suppose gn = 1. The
integral of this expression is called the “number staircase function” N (E)
and counts the number of levels (including their degeneracy) up to a given
energy E
Z
E
N (E) =
g (E) dE.
0
The density of states enters several important formulas and, as we will see
immediately, is related to the quantum propagator. First of all, a Laplacian
transform of g (E) yields the canonical partition function Z (β) :
Z ∞
X
e−βEn .
Z (β) = Lβ [g (E)] =
e−βE g (E) dE =
0
n
In order to relate g (E) to the propagator, let us consider initially the Fourier
0
transform of K with respect to the time difference t ≡ t − t:
Z
Z ∞
³ 0 ´
i
i ∞ ³ 0 ´ − i Et
i X ∗ ³ 0´
~
e ~ (E−En )t dt.
G q, q , E ≡ −
K q, q , t e
=−
ψn q ψn (q)
~ 0
~ n
0
Since the last integral oscillates and does not converge, we give a small
imaginary part to the energy: E → E + iε, so that
³ 0 ´ X ³ 0´
G q, q ; E =
ψn∗ q ψn (q)
n
1
.
E − En + iε
This Green function satisfies
³ 0 ´
³
´
0
(E − H) G q, q , E = δ q − q ,
(1.5)
1.2. THE QUANTUM PROPAGATOR
19
and for free theories it can be easily computed
µ
¶
¯
¯´
³ 0 ´
³
2m i
0¯
¯
D=1
G0 q, q , E = −
exp
−ik
q
−
q
¯
¯ ,
~2 2k
µ
¶
¯´
³ 0 ´
2m i + ³ ¯¯
0¯
D=2
G0 q, q , E = −
H
k
q
−
q
¯
¯ ,
~2 4 0
¯i
h ¯
µ
¶ exp ik ¯¯q − q 0 ¯¯
³ 0 ´
2m
D=3
G0 q, q , E = −
,
~2
4π |q − q 0 |
√
where k = 2mE. If we now take the trace of (26), i.e.
Z
³ 0 ´ X
1
b (E)
trG = dqG q, q , E =
≡G
E
−
E
+
iε
n
n
by using the identity
1
=P
x + iε
µ ¶
1
− iεπδ (x) ,
x
we see that
g (E) =
X
n
1
b (E + iε) = − 1
δ (E − En ) = − ImG
π
π
Z
³ 0
´
ImG q, q , E + iε dq.
From the above relation between g (E) and the canonical partition function, we also have that the density of states is given by the inverse Laplace
transform of the latter function
Z ε+i∞
1
g (E) = L−1
[Z
(β)]
=
eβE Z (β) dβ,
E
2πi ε−i∞
where the integral is to be taken in the complex β–plane along a contour
C which is parallel to the imaginary axis. The positive distance ε of the
contour from the imaginary axis has to be chosen large enough so that all
poles of the integrand lie to the left of the contour. If the integrand has
isolated poles we can easily evaluate this integral by the residue theorem.
20
CHAPTER 1. MATHEMATICAL INTRODUCTION
A very instructive example.
Let us consider the harmonic oscillator, with the spectrum given by
En = n (we will take care of the 12 term later). For such system the partition
function is easily done:
Z (β) =
∞
X
e−βn =
n=0
eβ/2
1
=
.
2 sinh β/2
1 − e−β
The poles of this function are all simple and all lying on the imaginary axis
at the values
βk = 2πik,
k = 0, ±1, ±2, ...
with residue (−1)k . By choosing the contour as in the figure
FIGURE
summing up all residues, and taking into account the extra factors exp (βk /2)
as well as exp (Eβk ) appearing in the inverse Laplacian transform, we have
g (E) =
X
+∞
X
δ (E − En ) =
n
e2πikE .
k=−∞
As we will see, this is equivalent to the Poisson resummation formula.
Since the spectrum is bounded by E = 0, only non–negative values of the
energy are relevant. The above formula can be written
)
(
∞
X
cos (2πkE) .
g (E) = 1 + 2
k=1
Note that the constant term, which corresponds to the average level density,
comes from the pole at β = 0, whereas the other poles along the imaginary
axis combine pairwise to make up all Fourier components of the oscillating
part.
¡
¢
It is now trivial to restore the actual dependence En = ~ω n + 12 of the
energy levels. In fact, we have
(
·
µ
¶¸)
∞
X
1
E
1
g (E) =
1+2
cos 2kπ
−
=
~ω
~ω 2
k=1
1
~ω
(
1+2
∞
X
k=1
µ
(−1)k cos
2πkE
~ω
¶)
.
1.2. THE QUANTUM PROPAGATOR
21
Comment.
Notice that g (E) naturally splits into a smooth part ge (E) and an oscillatory
part δg (E):
g (E) = ge (E) + δg (E) .
The smooth part is, in the above example, just the constant 1/~ω, namely
there is one energy level per unit ~ω. The oscillating part contains cosine functions with constant amplitudes; their arguments are multiples of
2πE/~ω. This quantity is easily recognized
p as the classical action: in fact,
taking the classical momentum p (x) = 2m (E − V (x)) and integrating
between the two turning points
p
2E/m
x0 = ±
,
ω
we obtain
I
S (E) =
Z
x
pdx = 2
p (x) dx =
−x0
2πE
.
ω
For the stair–case function the situation is as in the figure
FIGURE
The previous example is the simplest of a general theory. This states that
the smooth part is simply given by the corresponding integral on the classical
phase space:
Z
1
g (E) =
dq f dpf δ (E − H (q, p)) ,
(2π~)f
for a system of f degrees of freedom. This gives reason of the rule that the
number of eigenstates with energy lass than E equals the number of cells less
than E. Much more interesting is the oscillatory part, which is controlled
by the so–called Gutzwiller trace formula. It reads
·
¸
∞
X X
π
k
AΓk (E) cos SΓ (E) − σΓk
,
δg (E) =
~
4
Γ∈{ppo} k=1
where Γ counts classes of topologically distinct primitive periodic orbits
(ppo) (with energy E as a parameter); k counts the repeated revolution
around
each primitive orbit which yields a series of harmonics; SΓ (E) =
H
pΓ dqΓ is the classical action integral along the orbit Γ; and σΓk is the
”Maslov index ” of the orbit.
22
CHAPTER 1. MATHEMATICAL INTRODUCTION
Finally, the amplitudes AΓk depend on energy, time period and stability of
the orbits and their nature, being isolated or non–isolated. In the latter
case, one must sum over all distinct families of degenerate orbits.
The trace formula represents a Fourier decomposition of the oscillating part
of the level density. It gives the basis for many interesting interpretations of
quantum phenomena in terms of classical orbits. The reason why classical
periodic orbits are important can be understood by the path integral representation of the propagator. Suppose, in fact, that in the formula (1.3), we
put q0 = q and integrate on q, in order to obtain the trace of the propagator
FIGURE
In this case the action is a function of the variable q. The paths which
dominate the path integral are those for which pin = pf in , as it can be seen
from eq. (1.4). Those are just the periodic orbits
1.3
General spectrum depending on one quantum
number
It is simple to derive in full generality the previous result. Let us assume
that the spectrum En is given by a function f (n) and each level has a
degeneracy dn = D (n)
En = f (n) ;
dn = D (n) ,
n = 0, 1, ..
We assume f (n) to be a monotonic function with differentiable inverse
f −1 (x) ≡ F (x) so that
n = F (En ) .
Using the general properties of the δ..function, we have
¯
¯
δ (E − En ) = δ (E − f (n)) = δ (n − F (E)) ¯F 0 (E)¯ .
Further defining D (E) = D [F (E)], we have then
∞
¯
¯X
g (E) = D (E) ¯F 0 (E)¯
δ (n − F (E)) .
n=0
We can apply now the previous formula of the harmonic oscillator (i.e. Poisson resummation formula) and obtain
(
)
∞
X
¯ 0
¯
g (E) = D (E) ¯F (E)¯ 1 + 2
cos [2πkF (E)] .
k=1
1.3. SPECTRUM DEPENDING ON ONE QUANTUM NUMBER
23
This formula can be applied to any system where the values of En and
dn are known explicitly. This is not restricted to one–dimensional problems:
there exists in fact some high–dimensional potentials for which the spectrum
can be written analytically in terms of one single quantum number n with
known degeneracy. For all these systems the average level density is
¯
¯
ge (E) = D (E) ¯F 0 (E)¯ ,
whereas the oscillating part is
∞
¯ 0
¯X
¯
¯
cos [2πkF (E)] .
δg (E) = 2D (E) F (E)
k=1
Within the periodic orbit theory, 2π~F (E) is the classical action integrated
along an elementary closed orbit. The number k corresponds to the number
of revolutions around the orbit. If several distinct orbits exist, they have to
be summed separately. The amplitudes of the oscillation depend, in these
cases, on the chosen orbit and are, in general, much more difficult to obtain.
In fact the Bohr–Sommerfeld quantization rule reads
I
Z x2
S (E) = pdx = 2
p (x) dx = 2π~n
x1
(the eventual extra constant is irrelevant for the subsequent considerations).
Hence, since n = F (En ), we see by construction that
F (E) =
1
S (E) .
2π~
Example 1. Particle in a box of length L.
The exact spectrum is given in this case by
En = E0 (n + 1)2 ,
E0 =
q
In the above notation F (E) =
~2 π 2
,
2mL2
(n = 0, 1...)
E
E0
− 1, and therefore
(
"
r #)
∞
X
1
E
g (E) = √
1+2
cos 2πk
.
E0
2 E0 E
k=1
Example 2. Spherical harmonic oscillator in D–dimensions.
For this system the energy levels are given by one quantum number n with
a degeneracy dn which is a polynomial of order D − 1
24
CHAPTER 1. MATHEMATICAL INTRODUCTION
(Example
D=2
dn = n + 1
1
D=3
dn = (n + 1) (n + 2) .
2
Through the previous formulas we easily obtain
(
¶)
µ
∞
X
E
E
g (E) =
1+2
,
cos 2πk
~ω
(~ω)2
k=1
for D = 2, and
·
¸(
¶)
µ
∞
X
1
1
E
2
k
2
g (E) =
E − (~ω)
1+2
(−1) cos 2πk
4
~ω
2 (~ω)3
k=1
for D = 3.
1.4
The Gamma Function
In this Section, some of the basic properties of the Gamma Function will be
reviewed, due to its relevance in the subsequent considerations.
Γ (z) is an analytic function of the complex variable z, defined, for
Re(z) ≥ 0, by the following integral
Z ∞
Γ (z) =
dte−t tz−1 dt.
(1.6)
0
By making the change of variable t → t2 , it can be equivalently expressed
as
Z ∞
2
Γ (z) = 2
dte−t t2z−1 dt.
0
When z =
1
2 ,the
above is nothing but the gaussian integral. Therefore
µ ¶
√
1
Γ
= π.
2
The analytic continuation of Γ (z) can be achieved with many different techniques. First of all, it is easy to show that Γ (z) has simple poles at z =
0, −1, −2, ..., −n. Let us write (1) as
Z 1
Z ∞
−t z−1
Γ (z) =
dte t
+
dte−t tz−1 .
0
1
1.4. THE GAMMA FUNCTION
25
The second integral is convergent for all values of z in the complex plane.
About the first one, for Re(z) ≥ 0,where it converges uniformly, we can
expand the exponential and integrate term by term so that
Z
Γ (z) =
1
z−1
dtt
0
∞
X
(−1)k
k!
k=0
=
b (z) =
t +Γ
k
Z
∞
X
(−1)k
k=0
∞
X
(−1)k
k=0
k!
k!
1
b (z)
dttz+k−1 + Γ
0
1
b (z) .
+Γ
(z + k)
This expression clearly shows the poles at negative integer values of z.
Nearby the poles we have
Γ (z) '
(−1)k
.
k! (z + k)
Another way of extending analytically its definition in the complex plane is
by using its functional equation
Γ (z + 1) = zΓ (z) ,
(1.7)
which is immediately obtained by integrating by parts eq. (1). By virtue of
eq. (1.7), Γ (z) can be interpreted as a generalization of the factorial function
to complex numbers. Indeed, for integer positive values, we have
Γ (n) = (n − 1)!
Moreover, eq. (1.7) can be used to extend the definition of Γ (z) to other
values of the complex plane. Since Γ (z) = Γ(z+1)
, and Γ (z + 1) is defined
z
for Re(z) ≥ −1, we can use this relation to extend analytically the domain
of Γ (z) in the strip −1 ≤ Re(z) ≤ 0. Repeating the same argument, we can
extend Γ (z) to all values of the complex plane, except for z = 0, −1, −2.....
The Gamma function admits other interesting representations.
Infinite limit representation
Γ (z) = lim
n→∞
1 · 2 · 3 · .. · n
nz .
z (z + 1) (z + 2) ... (z + n)
Infinite product representation
z
∞ ³
Y
1
z ´e− n
γz
= ze
1+
.
Γ (z)
n
n=1
26
CHAPTER 1. MATHEMATICAL INTRODUCTION
To prove the first, let us introduce the function of two variables
¶
Z nµ
t n z−1
F (z, u) =
1−
t dt.
n
0
Since
µ
¶
t n
lim 1 −
= e−t ,
n→∞
n
we have
Z
∞
lim F (z, u) = F (z, ∞) =
n→∞
e−t tz−1 dt = Γ (z) .
(1.8)
0
We can evaluate F (z, u) in successive integrations by parts. Take v ≡
hence
Z 1
F (z, u) = nz
(1 − v)n v z−1 dv.
t
n,
0
Integrating by parts, we get
z
F (z, u)
n
n v
|1 +
=
(1
−
v)
nz
z 0 z
Z
1
(1 − v)n−1 v z dv.
0
The integrated part vanishes at both extremes. Repeating the procedure,
we have
Z 1
n (n − 1) ....1
F (z, u) = nz
v z+n−1 dv
z (z + 1) (z + 2) ... (z + n − 1) 0
=
1 · 2 · 3 · .. · n
nz
z (z + 1) (z + 2) ... (z + n)
and using eq. (1.8) we obtain Euler’s result.
The Weierstrass result is particularly interesting for our aim: in fact, it
resembles the spectral function for the Schrödinger problem! It can be easily
obtained from Euler’s limit form. Let us write it as
n
1 · 2 · ... · n
1 Y³
z ´−1 z
nz = lim
1+
n .
n→∞ z
n→∞ z(z + 1)...(z + n)
m
Γ (z) = lim
m=1
Inverting this expression and using the relation n−z = e−z ln n , we obtain
n ³
Y
1
z´
= z lim e−z ln n
.
1+
n→∞
Γ (z)
m
m=1
1.4. THE GAMMA FUNCTION
27
Multiplying and dividing by
·µ
¶ ¸
n
Y
z
1 1
1
exp 1 + + + ... +
z =
em
2 3
n
m=1
we obtain
1
= z lim exp
n→∞
Γ (z)
·µ
¶ ¸
n ³
Y
z ´ −z
1 1
1
1+
e m.
1 + + + ... + − ln n z · lim
n→∞
2 3
n
m
m=1
Since
Ã
lim
n→∞
∞
X
1
− ln n
k
!
=γ
k=1
we finally obtain the Weierstrass result
∞ ³
Y
1
z ´ −z
= zeγz
e n.
1+
Γ (z)
n
n=1
The Weierstrass definition shows that Γ (z) has simple poles at z = 0, −1, −2, ...
and that [Γ (z)]−1 has no poles in the finite complex plane.
This formula gives us the opportunity to discuss in general the (infinite)
product representation of entire functions. They generalize the concept of
polynomial expressions
P (z) = an z n + .... + a0 = A (z − z1 ) (z − z2 ) ... (z − zn )
which can be always decomposed in terms of the roots in the complex plane.
A function analytic for all finite z is called and entire function of z. The
logarithmic derivative f 0 /f is a meromorphic function with a pole expansion.
If f (z) has simple zeroes at z = an , we have
¸
∞ ·
f 0 (0) X 1
1
f 0 (z)
=
+
+
f (z)
f (0)
an zan
n=1
and integrating both sizes,
¶
¸
∞ · µ
f 0 (z)
f 0 (0) X
z
z
ln
=z
+
ln 1 −
+
.
f (z)
f (0)
an
an
n=1
Therefore, we arrive to the Weierstrass infinite product representation for
entire functions
· 0
¸ ∞ µ
¶
z
f (0) Y
z
f (z) = f (0) exp z
1−
e an .
f (0)
an
n=1
28
CHAPTER 1. MATHEMATICAL INTRODUCTION
The Gamma function satisfies a series of remarkable identities. One of these
is
Γ (z) Γ (1 − z) =
π
.
sin πz
(1.9)
To prove it, notice that, by virtue of the functional equation, the function
Φ (z) = Γ (z) Γ (1 − z) is periodic with period equal to 2,
Φ (z + 2) = Φ (z)
Moreover, it has simple poles at all points z ∈ Z. Hence by multiplying it
by
¶
∞ µ
Y
z2
z
1− 2
k
k=1
we obtain a function without singularities in the complex plane, i.e. a constant equal to 1, as it can be checked by taking the limit z → 0. But
¶
∞ µ
Y
z2
1
1 − 2 = sin πz
z
k
π
k=1
and eq. (1.9) is proved.
Another important result, due to Legendre, is the so called duplication formula, relevant in the theory of the Riemann zeta function.
µ
¶
√
1
Γ (z) Γ z +
= π21−2z Γ (2z) .
2
For future applications it is also useful to have the asymptotic expansion of
the Gamma function. It can be obtained by a saddle point approximation.
Performing the change of variable t = xz, we get
Z ∞
z+1
Γ (z + 1) = z
dxez(log x−x) .
0
The function ϕ (x) = ln x − x goes to −∞ both at x → 0 and x → ∞. It
presents a unique maximum at x = 1. For z → ∞, the above integral is
dominated by the region nearby this maximum. Expanding ϕ (x) in series
near this point
(x − 1)2
ϕ (x) = −1 −
+ ...
2
1.5. POISSON SUMMATION FORMULA
29
and substituting into (38) we have
Z ∞
√
(x−1)2
1
z+1 −z
Γ (z + 1) = z e
e−z 2 dx = 2πz z+ 2 e−z .
0
Hence
Γ (z + 1) =
√
1 −z
2πz z+ 2 e
µ
¶
µ ¶
1
1
1
ln Γ (z + 1) = z +
ln z − z + ln 2π + O
.
2
2
z
Let us mention also the integral representation
Z
¡
¢
e−η η z dη = e2πiz − 1 Γ (z + 1)
C
where the contour is shown in the Figure
FIGURE
This is particularly useful when z is not an integer; in this case the origin
is a branch point. The above equation can be easily verified by deforming
the contour as in Figure
FIGURE
The integral from ∞ to the origin gives −Γ (z + 1), placing the phase
of the logarithmic branch at the origin. The integral out of ∞ then yields
e2πiz Γ (z + 1), whereas the integral around the origin vanishes for z > −1.
It is simple to extend the range to include all non–integral values of z : the
integral exists in fact for z < −1, as far as we stay far from the origin.
Second, integrating by parts, one gets the usual functional equation
1.5
Poisson summation formula
Let us deal now with a very interesting application of Fourier transform:
the Poisson summation formula. Consider the function
S (x) =
+∞
X
n=−∞
f (x + nT )
30
CHAPTER 1. MATHEMATICAL INTRODUCTION
where f (x) is an arbitrary function. It is clear that S (x) is periodical with
period T :
S (x + T ) =
+∞
X
+∞
X
f (x + (n + 1) T ) =
n=−∞
f (x + nT ) = S (x) .
n=−∞
Therefore S (x) can be developed in Fourier series
+∞
X
S (x) =
Ck e
2πikx
T
k=−∞
where the coefficients are given by
1
Ck =
T
Z
T
2
− T2
S (x) e−
2πikx
T
dx.
By substituting the expression for S (x) we have
Z
1
Ck =
T
T
2
+∞
X
− T2 n=−∞
f (x + nT ) e−
2πikx
T
dx.
Making the change of variable y = x + nT , we get
Ck =
Z
+∞ Z n+ T
2
2πiky
2πiky
1 X
1 +∞
f (y) e− T dy =
f (y) e− T dy.
T n=−∞ n− T
T −∞
2
The last term is the Fourier transform of the function f (x) (let us denote it
by fb(ω)), and then
µ
¶
2π b 2πk
f
.
Ck =
T
T
Finally, the function S (x) can be rewritten as
µ
¶
+∞
2πikx
2π X b 2πk
S (x) =
f (x + nT ) =
f
e T .
T
T
n=−∞
+∞
X
k=−∞
This is the Poisson Summation Formula. Let us see some applications.
Example 1
Compute
S0 =
+∞
X
n=0
a2
1
.
+ n2
1.5. POISSON SUMMATION FORMULA
31
This series can be written as
+∞
1
1
1 X
S0 = 2 +
.
2a
2 n=−∞ a2 + n2
Let us apply the PRF to the second term, where
f (v) =
a2
1
,
+ v2
T = 1,
x = 0.
The Fourier transform of f (v) is
Z +∞
1
1
b
e−ivν dv.
f (v) =
2
2π −∞ a + v 2
This integral can be computed using the residue theorem.
a) v < 0. Close the contour as in figure..
The integral on the semicircle vanishes when the radius goes to infinity, so
it remains the contribution of the residue:
¸
·
1 −a|v|
1
−ivν
e
=
e
.
I = iRes 2
a + v2
2a
v=ia
b) v > 0. we get again the same result. Consequently
+∞
1
1
π
1
π X −a|2πk|
π
e
= 2−
S0 = 2 +
+
.
2a
2a
2a
2a a 1 − e−2πa
k=−∞
Let us prove as an example the following identity
∞
X
1
π2
.
=
n2
6
n=1
Notice that
S0 =
∞
X
n=0
therefore
∞
X
1
1
1
=
+
,
2
2
2
2
a +n
a
a + n2
n=1
·
¸
∞
X
1
1
= lim S0 − 2 .
n2 a→0
a
n=1
Expanding in series S0
2
(2πa)
1
π
π (1 + 2πa + 2 + ...)
S0 ' 2 −
+
=
2a
2a a 2πa(1 + 2πa + (2πa)2 + ...)
2!
3!
32
CHAPTER 1. MATHEMATICAL INTRODUCTION
µ
¶
1
π
1
(2πa)2
2πa (2πa)2
2πa 2
−
+
(1 + 2πa +
+ ...)(1 −
−
+
...) =
2a2 2a 2a2
2
2!
3!
2!
·
¸
π
1
π
1 2 4π 2
π2
1
1
−
+
+
+
π
−
=
+ 2,
2
2
2a
2a 2a
2a 2
6
6
a
hence the result. The Poisson resummation formula is clearly useful from
the point of view of numerical calculus.
Example 2
Consider the series
S (t) =
+∞
X
2
e−k t .
−∞
Suppose that we want compute it at t = 0.01. In this case, to have a
precision of 10−10 we need ∼ 50 terms. Using Poisson formula we have
r
S (t) =
+∞
π X − π2 n2
e t .
t −∞
At t = 0.01, we get
r
S=
´
π ³
2
1 + 2e−100π + ... = . . .
0.01
Exercise 1
Show that
S1 =
Exercise 2
Show that
∞
X
1
π 1 − e−πa
=
.
4a 1 + e−πa
a2 + (2n + 1)2
n=0
∞
X
(−1)n
1
π e−πa
S2 =
=
+
a2 + n2
2a2
a 1 + e−2πa
n=0
Exercise 3
Prove that
∞
X
(−1)n
n=0
n4
=−
7π 4
.
720
1.6. SPECTRAL FUNCTIONS
1.6
33
Spectral Functions
The previously introduced functions are some representatives of a large class
of functions which deal with the spectrum of a system. They will be the
object of our future study. To fix the ideas, let us take a one–dimensional
system with Hamiltonian
H=
p2
+ V (q) ,
2m
[p, q] = i~
(1.10)
with the eigenvalues which goes like
En ∼
= nα
(1.11)
(we will see later how to compute these quantities). Let us define the following spectral functions
• Partition function
∞
X
Z (t) = T r exp(−tH) =
e−tEn ,
Ret ≥ 0
n=0
• Resolvent Trace
−1
R (E) = T r (H + E)
=
∞
X
1
,
E + En
n=0
E ∈ C− {−}
(1.12)
• Fredholm determinant
¶
∞ µ
¡
¢ Y
E
−1
∆ (E) = det 1 − EH
=
1−
,
En
E ∈ C.
(1.13)
1
.
a
(1.14)
n=0
• Zeta function,
ξ (s) = T rH
−s
∞
X
1
=
,
Ens
Res >
n=0
• Mellin Transform of R (E)
Z
M (s) =
0
∞
E −s R(E)dE
(1.15)
34
CHAPTER 1. MATHEMATICAL INTRODUCTION
Few comments
(i) Note that we have defined the resolvent trace with (−E), comparing with
the previous formula.
(ii) In the Fredholm determinant we assume the absence of Hadamard factors, i.e. we assume
X 1
< ∞.
En
This formula can be modified consequently if this condition of convergence
is not satisfied.
It is easy to see that all these functions are related to each other, a result
which is intuitive, since all of these are based on the spectrum {En }. We
have
Z
∞
R (E) =
Z (t) e−tE dt
0
in the region where Re E > 0. Similarly
¶
·Z E
¸
∞ µ
Y
E
∆ (−E) =
1+
= exp
n (λ) dλ ,
En
0
E ∈ C.
(1.16)
n=0
Moreover,
Z
∞
Z (t) t
s−1
dt =
∞ Z
X
∞
e−tEn ts
n=0 0
0
dt
t
making the change of variables tEn = x we have
Z ∞
X 1 Z ∞
s−1
Z (t) t dt =
dxe−x xs−1 = Γ (s) ξ (s) .
s
E
0
n 0
Finally
ξ (s) =
1
sin πsM (s) ,
π
(1.17)
which can be proved as follows.
XZ ∞
XZ ∞
1
1
dE =
dEe−s ln E
.
E −s
E
+
E
E
+
E
n
n
0
0
n
n
In order to compute this integral, let us choose the contour as in the figure
FIGURE
1.6. SPECTRAL FUNCTIONS
35
In going to the lower branch II, the logarithmic function changes, since
E → Ee2πi . Therefore, applying the theorem of residues to the above contour, with vanishing contribution from the external circle and the one near
the origin, we have
X
¡
¢
I 1 − e−2πis = 2πie−iπs
En−s .
n
Hence we get formula (1.17).
36
CHAPTER 1. MATHEMATICAL INTRODUCTION
Chapter 2
Semiclassical methods:WKB
Approximation
2.1
Introduction
Consider the one dimensional Schrödinger equation
d2 Ψ 2m
+ 2 [E − V (x)] Ψ = 0.
dx2
~
(2.1)
If the potential doesn’t have a simple form, solving the Schrödinger equation
is usually a complicated problem, which requires the use of approximate or
numerical methods, such as
1) Perturbation theory
2) Variational methods
3) Numerical diagonalization.
One particular method has found its greatest use in the case of one–dimensional
systems. This is the so called WKB method (after Wentzel, Kramers and
Brillouin).
The basic idea is very simple and it is similar to the eikonal expansion in
optics.
Notice that if V = const, eq. (2.1) has the solutions Ψ± = e±ikx . This
suggests to look for a solution of the form
Ψ = eiS(x) .
Substituting into eq. (2.1), we get
00
iS − (S 0 )2 + [K (x)]2 = 0,
37
38
where
CHAPTER 2. WKB APPROXIMATION
 q
2m

~ [E − V (x)],
q
K (x) =
 −i 2m [V (x) − E]
~
E > V (x)
E < V (x)
The above equation can be solved recursively, i.e. substituting the n–th
approximation, for the (n + a)–th approximation we have
Z xp
K 2 (x) + iSn00 (x) + Cn+1 ,
Sn+1 = ±
hence
Z
S0 = ±
Z
S1 = ±
x
K (x) + C0 ,
xp
K 2 (x) ± iK 0 (x)dx + C1 .
Obviously, to make sense of this procedure, the next approximation should
be close to the previous one, which for the above terms is equivalent to say
¯ 0
¯
¯
¯
¯K (x)¯ ¿ ¯K 2 (x)¯
(2.2)
In the above equations, both signs have to be chosen equal, since S1 is supposed to be an improvement of S0 . Hence
r
¸
Z x
Z x·
K0
K0
S1 = ±
K (x) 1 ± i 2 dx ' ±
K (x) ± i
dx =
K
K
Z x
i
±
K (x) + log K (x) + C.
2
Therefore, at this level of approximation, for the wave function we have
· Z x
¸
A
Ψ (x) ' p
exp ±i
K (t) dt .
K (x)
The condition (2.2) has a very physical meaning. We can define, in the
regions where E > V (x), an effective wave length
λ (x) =
2π
k (x)
and therefore eq. (2.2) can be written
¯ ¯
¯ dp ¯
λ (x) ¯¯ ¯¯ ¿ |p (x)| ,
dx
2.2. THE CONNECTION FORMULAS
39
where p (x) = ±~K (x) is the momentum of the particle. Hence, the range
of the validity of the WKB approximation (at this order) is that the change
of the momentum over a wavelength must be small compared to the square
of its amplitude.
The method falls down if K (x) vanishes or if K (x) varies very rapidly.
This certainly happens at the classical turning point for which
V (x) = E,
or whenever V (x) has a very steep behaviour. In both cases a more accurate
solution has to be found. Obviously, the WKB method would not be useful
unless we find a way to analyze such situations.
2.2
The connection formulas
Consider a potential as in the figure
FIGURE
Let us consider initially the case x À x2 . In this region the classical momentum is imaginary and therefore one of the solutions blows up. Hence we
have to consider only the exponentially decaying part, i.e.
· Z x
¸
¯ 0
¯
c
¯K (x)¯ dx0 .
ψ (x) = p
exp −
(2.3)
|K (x)|
In the region x1 < x < x2 , on the other hand, far away from the turning
points, we have
· Z x ³ ´
¸
· Z x ³ ´
¸
A
B
0
0
0
0
ψ (x) = p
exp i
K x dx + p
exp −i
K x dx
K (x)
K (x)
(2.4)
Let us consider now more closely the situation near one of the turning point,
let’s say x = x2 . Suppose that the potential is smooth and admits a series
expansion. Keeping only the leading term we have
d2 Ψ
2m
0
= 2 (x − x2 ) V (x2 ) Ψ.
dx2
~
By posing
2m 0
V (x2 ) ≡ α3 ,
~2
y = αx,
40
CHAPTER 2. WKB APPROXIMATION
the above equation becomes
d2 Ψ
= y Ψ.
dy 2
This is the differential equation of the Airy function (see later). Disregarding
once again the exponential growing function, its solution is expressed by the
so called Airy function Ai (y),
defined by
µ 3
¶
Z ∞
1
v
Ai (y) = √
dv cos
+ vy .
3
π 0
For negative y, corresponding to the interior of the potential, it is oscillatory,
whereas for positive y is exponentially decaying. Its asymptotic expansions
are

³
´
1
3
 |y|− 4 sin 2 |y| 2 + π ,
y → −∞
³3
´4
Ai (y) =
.
1
3
−
1
2
 y 4 exp − y 2 ,
y → −∞
2
3
We have to match this behaviour with our previous expressions. We have
π
B = −A = iCei 4 ,
and therefore for x1 < x < x2 we have now
· Z x2 ³ ´
¸
2C
π
0
0
Ψ= p
sin i
.
K x dx +
4
K (x)
x
(2.5)
Inspecting the expressions (2.3) and (2.4), we see that a phase factor π4 has
been introduced in the wave function at the turning point of a smooth
potential.
The WKB condition can be now obtained by demanding single–valuedness of
the wave function. In fact, repeating now the analysis for the other turning
point, we arrive to the expression
· Z x2 ³ ´
¸
e
2C
π
0
0
Ψ= p
sin i
.
(2.6)
K x dx +
4
K (x)
x
The two phase factors in eqs. (2.5) and (2.6) should be the same up to
a negative sign, which can be absorbed in the overall function. So, with
e = (−1)n C, we have
C
Z x ³ ´
Z x2 ³ ´
π
π
0
0
0
0
K x dx + = −
K x dx − + (n − 1) π,
4
4
x1
x1
2.2. THE CONNECTION FORMULAS
i.e.
Z
x2
x1
µ
¶
1
K (x) dx = n +
π,
2
41
n = 0, 1, 2, ...
By extracting the ~ dependence, this becomes the familiar equation of quantization
µ
¶
I
Z x2
1
p (x) dx = p (x) dx = n +
2
h.
2
x1
The left–hand side of this equation is equal to the area enclosed by the curve
representing the motion in phase space and it is called the phase integral I.
On the other hand, it also measure the phase change which the oscillatory
wave function Ψ (x) undergoes between the turning points.
Z
x2
K (x) dx =
x1
µ
¶
1
h.
n+
2
Dividing this expression by 2π, we see that according to the WKB approximation
n 1
+
2 4
of quasi wavelengths fit between x1 and x2 . Hence, n represents the number
of modes of the wave function, a fact which helps in visualizing the elusive
ψ. According to eq. (23), the area of the phase space between one bound
state and the next one is equal to h. From this observation we see that it
comes the statement often used in statistical mechanics, that each quantum
state occupies a volume h in phase space.
In order to find the normalization of the semi–classical wave function, we
assume that the contribution beyond the turning points is negligible, so
Z
A2u
hence
x2
x1
dx
sin2
p (x)
µZ
x
x1
³ 0 ´ 0 π ¶ A2 Z x2 dx
K x dx +
'
,
4
2 x1 p (x)
2
A2 = R x2 dx .
x1 p(x)
With all the above quantities fixed, it is nice to check how the semi–classical
wave functions match the exact ones. This check can be done easily for
the harmonic oscillator. Obviously at the turning points the wave function
blows up and one need the connecting formulas.
42
2.3
CHAPTER 2. WKB APPROXIMATION
Infinite well
The quantization condition changes if one of the walls of the potential is an
infinite step. In this case the wave function is zero at the boundary. Suppose
then that we have two turning points, one x2 with a smooth behaviour and
x1 with an infinite well. In this case, coming from right hand side of x2 , we
have for the wave function
·Z x 2
¸
π
Ψ ∼ sin
K (x) +
.
4
x
But then we have to impose
Ψ (xc ) = 0.
Therefore we have the quantization condition
Z x2
π
K (x) dx + = (n + 1) π
4
x1
and therefore, in this case
I
µ
¶
3
p (x) dx = n +
h.
4
Since π4 of the phase of the right hand side of the above equation may be
attributed as before to the turning point x2 , the extra phase factor π/2 has
to be attributed to the step well.
2.4
Airy function
Consider the Schrödinger equation in a linear potential V = Ax,
d2 Ψ
2m
= 2 [Ax − E] Ψ.
dx2
~
It is obvious that if we shift the variable
x→x+
E
,
A
we can absorb the eigenvalue E, so if we find a solution for one value, we
find it for all other values. We write then
2m
d2 Ψ
= 2 AxΨ.
2
dx
~
2.4. AIRY FUNCTION
43
Introduce the quantities
µ
2mA
α =
~2
y = αx.
¶
3
,
The above equation becomes
d2 Ψ
= yΨ.
dy 2
The Fourier transform of Ψ is
Z
+∞
Ψ (y) =
dpΨ (p) eipy .
−∞
Then we have the differential equation
i
whose solution is
·
¸
p3
Ψ (p) = exp −i
.
3
Hence
Z
+∞
Ψ (y) =
−∞
with
d
Ψ (p) = p2 Ψ,
dp
¶¸
· µ
√
p3
= 2 πΦ (y) ,
dp exp i py −
3
1
Φ (y) = √
π
Z
∞
0
µ
¶
p3
cos yp +
dp.
3
This function can be written in terms of Bessel functions
 q
³ 3´
y
2 2

K
,
y>0
1
3π h3 3³y
´
³
´i
Φ (y) =
.
3
√
 1 πy J 1 2 y 2 + J 1 2 y 23 ,
y
<
0
−
3
3
3
3
3
Hence, it goes exponentially to zero for y → +∞, whereas has an infinite
number of zeros for y < 0. [Comments. Ideal spectral function! Up to
Hadamard factors,
¶
Yµ
E
∆ (E) =
1+
,
λi
i
with λi zeros of Airy function.
44
2.5
CHAPTER 2. WKB APPROXIMATION
WKB for radial motion
The previous considerations can be easily extend to deal with quantum
mechanical systems in a central potential. Let us consider first a two–
dimensional problem
¸
·
1 ∂
~2 ∂ 2
1 ∂2
+
Ψ (r, ϕ) + V (r) Ψ (r, ϕ) = EΨ (r, ϕ) .
−
+
2m ∂r2 r ∂r r2 ∂ϕ2
The radial and angular eqs. are separable in this case. Furthermore, by
√
dividing the radial wave function by r, we obtain for it an eq. in the
radial variable which is similar to the one–dimensional case, i.e.
Ψ (r, ϕ) =
X vl (r)
√ eilϕ
r
l
and
Q2l (r) =
¶¸
·
µ
2m
~2
1
2
,
E
−
V
(r)
−
l
−
~2
2mr2
4
where ~l is the eigenvalue of the angular momentum. The above formulas
are indeed similar to the one dimensional case and we are tempted to write
down analogous WKB expressions, with
p (x)
→ Ql (x) .
~
To check if this procedure gives the correct asymptotic phase in the wave
function, let’s analyze a solvable case.
2.6
Square–well potential
Consider V (r) to be a square–well potential of depth V0 Let us also assume
l 6= 0. The exact regular solution is Jl (kr), where Jl is the Bessel function
and the wave number
2m (E − V0 )
k2 =
~2
is a constant. For large r, such that
kr À |l| ,
r
Jl (kr) '
µ
¶
2
lπ π
cos kr −
−
.
πkr
2
4
2.6. SQUARE–WELL POTENTIAL
45
Let us construct the WKB wave function for this
To match the
¡ 2problem.
¢
1
2
asymptotic form let’s do in Ql the replacement l − 4 → s2 , where s2 is
a parameter to be determined shortly.
Then, using the previous formulas, the WKB formula for vl is
·Z r
¸
¡ 0¢ 0 π
2
e
q
cos
Qc r dr −
vl (r) '
4
r1
fl (r)
Q
where
·
¸
2m
~2 2
e
Qc (r) ≡ 2 E − V0 −
s
~
2mr2
and r1 is the turning point near the origin determined by the condition
k 2 r12 − s2 = 0.
The integral in (10) can be done explicitly
Z rh
q
i1/2 dr0
³s´
2
2 02
2
2 − s arccos
I=
k r −s
=
(kr)
−
s
.
r0
kr
r1
In the asymptotic region kr À s, it is equal to
π
I ' kr − s .
2
Comparing now with (8), we see that
s2 = er ,
in order to have a match of the phases. In turns out that this prescription
is also valid in the more general case where V (r) varies with r, i.e. we have
to use the substitution
1
l2 − → h2 .
4
This condition can be justified by the request that the angular part of the
wave function is semi–classical as well.
We can now derive the WKB quantization condition for the bound states.
Define initially the phases
Z r ³ ´
e l r0 dr0
φl (r1 , r) ≡
Q
r1
Z
φl (r1 , r2 ) ≡
r2
r1
³ ´
e l r0 dr0 .
Q
46
CHAPTER 2. WKB APPROXIMATION
Since the WKB wave function at r may be written in terms of either one of
them, it follows that
³
³
π´
π´
cos φl (r1 , r) −
= ± cos φl (r1 , r2 ) −
,
4
4
and therefore we arrive to the quantization condition
µ
¶
Z r2
1
e
Ql (r) dr = nl +
φl (r1 , r2 ) =
π.
2
r
Defining the action variable
Ir ≡
1
1
Sr =
2π
2π
I
pr dr,
the quantization condition may be written in the generalized form
³
µ´
Ir = nr +
~,
4
where µ is called the Maslov index. In the above example µ counts the
number of classical turning points, where the amplitude of the wave function
diverges.
Note that in this
the WKB wave function over a complete cycle acquires
¡ case
µ¢
a factor ±2π nr + 4 .
In case the particle encounters a hard wall b times, the wave function goes
to zero under Dirichel b.c. at every encounter and picks an extra phase bπ
for the b reflections. The total acquired phase may be written then as
¶
µ
µ b
.
2π nr + +
4 2
Under Neumann b.c., on the other hand, there is no change in the phase of
the wave at the wall and b = 0, in this case. Thus the above quantization
condition may be written as
µ
¶
µ b
I r = nr + +
~.
4 2
This formula is important for obtaining Gutzwiller trace formula in the
integrable cases, although it does not apply for the non–integrable ones.
Finally, in the 3–dimensional case, the semiclassical formalism is set up by
the substitution
¶
µ
1 r
,
l (l + 1) → l +
2
obtained by the same request of semi–classical behaviour for the angular
wave function.
2.7. CONNECTION FORMULAS
2.7
47
Connection formulas for semiclassical wave functions
The previous analysis can be done in another way which is very instructive.
This provides the connection formula between semi–classical wave functions
in the allowed and forbidden classical region. Let us consider first a turning
point in the right part, as in figure
FIGURE
In the region I : x > a, far away from the turning point we have
·
¸
Z
C
1 x
Ψ = p exp −
|p| dx
~ a
|p|
(2.7)
whereas, in the region II : x < a, the wave function is given by a linear
superposition of the two allowed solutions
· Z x
·
¸
¸
Z
i
i x
A
B
Ψ = √ exp
pdx + √ exp −
pdx .
(2.8)
p
~ a
p
~ a
How can we determine the coefficients A and B ?
In the vicinity of the turning point we have
E − V (x) ' F0 (x − a) ,
F0 = −
dv
|x=a < 0.
dx
Suppose now that we can still use the semiclassical approximation sufficiently close to the turning point. In this case, in the region I we have
Ψ= p
C
1
2m |F0 | (x − a)1/4
·
¸
Z
√
1 xp
exp −
2m |F | x − adx .
~ a
(2.9)
Consider now this function as a function of the complex variable x and the
passage from positive to negative (x − a) must be along a path which is
always sufficiently far from the point x = a, so that the WKB is still valid.
Let us consider first the variation of this function from right to left along a
semicircle of radius ρ in the upper half–plane.
FIGURE
48
CHAPTER 2. WKB APPROXIMATION
Along this semicircle
µ
¶
3
3
x − a = ρe ,
cos ϕ + i sin ϕ .
2
2
a
¢
¡
The exponential factor in (2.9), at the beginning 0 < ϕ < 23 π increases in
modulus and then decreases to modulus 1. At the end of the semicircle the
exponent becomes purely imaginary and equal to
Z
Z
i xp
i x
−
2m |F0 | (a − x)dx = −
p (x) dx.
~ a
~ a
Z
iϕ
x√
2 3
2 3
t − adt = z 2 = ρ 2
3
3
In the coefficient of the exponential, the change along the semicircle is
1
1
π
(x − a)− 4 → (a − x)− 4 e−i 4 .
Hence the function (2.7) has become the second term in (2.8) with
π
B = Ce−i 4 .
Why is it possible to determine only the coefficient B passing in the upper
complex plane? There is a simple explanation of this fact. Suppose we
follow the variation of the function (2.8) along the same semicircle but in
the opposite direction.
FIGURE
In this case we have, as before
µ
¶
Z xp
3
2 3
3
(a − x)dx = ρ 2 cos ϕ + i sin ϕ .
3
2
2
a
Correspondingly, for the exponentials
·
¸
·
¸
2 2
3
i2 2
3
3
3
A exp − ρ sin ϕ exp
ρ cos ρ +
3
2
~3
2
·
¸
·
¸
2 2
3
i2 2
3
B exp ρ 3 sin ϕ exp − ρ 3 cos ρ .
3
2
~3
2
We see that at the very beginning of the path the first term becomes exponentially small with respect to the second and it must be discarded! So, in
this way we have lost the information on A. In order to obtain A, we must
continue the function (2.7) along a path in the lower half plane.
2.7. CONNECTION FORMULAS
49
FIGURE
Repeating the exercise we find that along this path the exponential has
become the one of the term in A, whereas its prefactor has become
1
π
(x − a)− 4 → (a − x) ei 4 .
In this way we find
π
A = Cei 4
and therefore the final solution is
( C
£
¤
Rx
√ exp − 1
|p| dx ,
p
~
a
£1 R x
¤
Ψ=
2C
π
√
p cos ~ a pdx + 4 ,
x>a
x<a
¯
¯
¯
¯.
¯
What has it been the connection formula for the divergent solution? Namely,
suppose that in I we have
· Z x
¸
C
1
Ψ = √ exp
|p| dx .
p
~ a
What is the analytic continuation in x < a ? Repeating the calculations,
coming from the path in the upper plane, we find this time
π
A = Ce−i 4 ,
whereas, along the path in the lower plane
π
B = Cei 4 .
Hence
· Z x
·
· Z x
¸
¸
¸
Z
i
i x
C
1
C −i π
C iπ
π
π
2
2
exp −
=
|p| dx → √ e exp
pdx + i + √ e
pdx − i
√ exp
p
~ a
p
~ a
4
p
~ a
4
·Z x
¸
1
π
− √ C sin
pdx +
.
p
4
a
So, we have the connection formulas
· Z x
¸
·
¸
Z
1
1
π
1
i x
cos
pdx
+
→
−
exp
−
|p|
dx
,
√
√
p
~ a
4
p
~ a
· Z x
¸
· Z x
¸
1
1
π
1
1
pdx +
→ − √ exp
|p| dx .
√ sin
p
~ a
4
p
~ a
£ Rx
¤
From these formulas we see that an “innocent” component sin ~1 a pdx + π4
in the allowed region produces a dangerous component outside!
50
CHAPTER 2. WKB APPROXIMATION
2.8
Transmission through a barrier
Suppose we have a barrier potential with a particle hitting from the left the
barrier, with an energy E insufficient (classically) to pass on the right. We
would like to compute semiclassically the transmission coefficient.
For the semiclassical solution, in the three regions, we have
£ Rx
¤
£ Rx
¤
 A
¯
B
¯
√
√
x<a

 p exp £i Ra kdx +¤ p exp −i£R a kdx ,¤
¯
x
x
¯
C
C
√ exp −
√ exp
|k|
dx
+
|k|
dx
,
a
<
x
<
b
Ψ=
¯.
p
p £
a
a
£
¤
¤
R
R

¯
 √F exp i x kdx + √G exp −i x kdx ,
¯
x>b
p
p
b
b
Using the previous connection formula, we can easily establish the linear
combination between the coefficients and the final result is very simple.
¡
¢ ¶µ
µ
¶
µ
¶
1
1
1 2θ +
A
i
2θ
−
F
2θ
2θ
¡
¢
=
,
1
1
B
2θ + 2θ
G
2 −i 2θ − 2θ
where
Z
θ ≡ exp
b
|p| dx.
a
This parameter measures the height and the thickness of the barrier. For
the transmission coefficient, we have simply (assuming G = 0, i.e. no wave
from the right)
¯ ¯2
¯F ¯
4
T = ¯¯ ¯¯ = ¡
¢ .
1 2
A
2θ + 2θ
For a high and broad barrier θ À 1 and
¸
·
Z
1
2 b
T ' 2 = exp −
|p| dx .
θ
~ a
2.9
Metastable states
Consider now the same problem of computing the transmission amplitude
for a potential as in the figure
FIGURE
i.e., a symmetric potential with respect to the origin and a deep well inside.
Repeating the previous calculations, one obtains
4
¢
T =¡ 2
,
1
4θ + 4θ2 cos2 L + sin2 L
2.10. POTENTIAL V (X) = λ |X|A
where
51
Z
θ ≡ exp
b
|k| dx,
a
Z
a
L≡
k (x) dx.
−a
The above quantity reaches its maximum when cos L = 0, i.e. when
µ
¶
1
L=
K (x) dx = n +
π,
2
−a
Z
a
the quantization condition for the bound states!
FIGURE
This is a very important observation. In fact, suppose we want to determine
if a given number belongs or not to a given sequence {En }. If we are able
to find a potential which has {En } as a spectrum, the above question, of
a mathematical nature, becomes a physical question. In fact, once such
potential is constructed, we round it to make its eigenstates metastable
and we perform a scattering experiment, with a beam of particles hitting
the target with the energy equal to the number we want to probe. If the
particle passes with probability 1, we have hit an eigenvalue! Obviously,
once we round the potential, its eigenvalues acquire an imaginary part. In
fact, there is now a probability to escape from the well and to go at infinity.
We have a decay process
1
En → En − iΓn ,
2
with
Γn ≡
2.10
θ2
1
¡ ∂L ¢
.
∂E E=En
Potential V (x) = λ |x|a
Consider the Schrödinger equation in one variable:
−
~2 d2
Ψ (x) + V (x) Ψ (x) = EΨ (x) .
2m dx2
52
CHAPTER 2. WKB APPROXIMATION
We want to write initially the Schrödinger equation in homogeneous variable.
To this aim, let us introduce the dimensionless quantity
η=
x
,
ξ
where ξ is a length scale. We have
·
¸
~2 d2
a
a
−
+
λξ
|η|
ψ = Eψ.
2mξ 2 dη 2
Hence
·
¸
2
b0 − 1 d + b |η|a ψ = Eψ,
E
2 dη 2
where
2
b0 = ~ ,
E
mξ 2
b=
mλ a+2
ξ .
~2
(2.10)
b0 . Notice that b is
From now on, all the energies will be measured in units of E
a dimensionless parameter. It will be fixed through the WKB quantization.
2.10.1
WKB quantization of V (x) = λ |x|a
Consider the following potential.
FIGURE
We have:
I
¶
µ
1
h.
p (x) dx = n +
2
For the left–hand side we have
I
Z xp
Z
p (x) dx = 4
2m (E − V (x))dx = 4
0
xq
2m (E − λ |x|a )dx =
0
Z
√
1/2
2m4E 4
x
r
1−
0
Making the change of variable y a =
³
λ1/a x
E 1/a
λ |x|a
dx.
E
´a
, the above integral becomes
√ E 12 + a1 Z 1 p
4 2m 1/a
1 − y a dy.
λ
0
2.11. SEMICLASSICAL ESTIMATE OF hX α iN
Since
Z
I (a) =
1p
1−
y a dy
0
53
¡
¢
√
π Γ 1 + a1
¡
¢,
=
2 Γ 32 + a1
we finally have for the LHS
¡
¢ a+2
I
√
Γ 1 + a1 E 2a
p (x) dx = 2 2πm ¡ 3 1 ¢ 1 .
Γ 2 + a λa
Equating this formula to the RHS, we deduce the quantization of energy
levels:
µ
¶ 2a
1 a+2
En = E0 n +
,
2
"
# 2a
¡
¢
~ λ1/a Γ 32 + a1 √ a+2
¡
¢
E0 = √
π
.
2m Γ 1 + a1
In order to compare the two results we have to make equal the two energy
b0 . One way of doing this consists in extracting initially
scales, i.e. E0 and E
ξ from the second of (2.10)
µ
ξ=
~2 b
mλ
1
¶ a+2
,
b0 , which becomes
and then substituting it into E
" √
# 2a
1
a+2
a
~
2
λ
b0 = √
E
.
2m b a1
Comparing now this expression with the corresponding one for the semiclassical expression, the two ones match if we choose
"r
¢ #a
¡
1
Γ
1
+
2
a
b0 =
¢ .
¡
E
π Γ 32 + a1
2.11
Harmonic oscillator: semiclassical estimate
of hxα in
There is a beautiful application of the semiclassical formula. This consists
in the semiclassical estimate of
Z +∞
α
hn| x |ni ≡
dxxα ψn (x) ψn (x) dx,
−∞
54
CHAPTER 2. WKB APPROXIMATION
where
ψn (x) =
µ ¶1
1 4 1 1 − x2
√ e 2 Hn (x) .
π
2n/2 n!
The difficulty of the problem consists in the lacking of suitable approximation of the Hermite polynomials with respect to the variable n. We
can get around this problem by using the semiclassical approximation. Before doing that, we have to fix a convenient notation. In the units where
m = ω = ~ = 1, the exact energy levels (and the semiclassical ones) are
En = n + 12 . The position of the inversion points an
FIGURE
is given by the condition
V (an ) = En ,
i.e.
1 2
1
a =n+ ,
2 n
2
and therefore
an =
√
2n + 1.
The momentum p (x) is given by
p
p
p (x) = 2m (E − V (x)) = a2n − x2 .
With the above notation the semiclassical wave function in the allowed region −a < x < a (|x| > a is supposed to give a negligible contribution to all
the following computation) is represented by
·Z x
¸
π
An
p (t) dt −
.
ψn (x) ' √ cos
p
4
−a
The normalization constant An is easily fixed. In fact
·Z x
¸
Z a
Z a
Z
π
A2 a dx
dx
2
2
2
cos
p (t) dt −
'
.
ψn dx = An
4
2 −a p (x)
−a
−a
−a p (x)
Here we have taken the average 12 of the rapid oscillating trigonometric function. Since
Z a
Z a
Z 1
dx
dx
dt
√
√
=2
=2
= π,
2
2
a −x
1 − t2
−a p (x)
−a
0
2.11. SEMICLASSICAL ESTIMATE OF hX α iN
55
for the normalization constant A we have
r
2
A=
.
π
Last observation. We can write the wave function as
"
#
Z x/an p
π
A
1 − t2 dt −
ψn =
cos 2a2n
,
·
³ ´2 ¸ 14
4
1
−1
an2 1 − axn
i.e.
ψn =
£
¤
cos a2n f (η) ,
A
1
2
1
4
an [1 − η 2 ]
where η ≡ axn . By using the above formula is now easy to compute the
average of xα . We have
Z an
Z
£
¤
A2 an α
1
hxα i =
xα ψn (x) ψn (x) dx =
x p
cos2 a2n f (η) dx.
an −an
1 − η2
−an
By making the change of variable x → x/an and taking the average of the
rapid oscillating function, we have
Z
α
hx i =
A2 aαn
0
1
tα
dt √
.
1 − t2
For the last integral we obtain
Z
I (α) ≡
0
i.e.
1
¡
¢
√
π Γ 1+α
tα
¡2¢ ,
dt √
=
α Γ α2
1 − t2
¡
¢
α
2 1 Γ 1+α
2
¡ α ¢ (2n + 1) 2 .
hn| x |ni = √
πα Γ 2
α
Simple checks of the above formula are obtained by taking
α=2
¡ ¢
­ 2®
1
1
2 1 Γ 32
¡ 1 ¢ (2n + 1) = (2n + 1) = n + ,
x =√
2
2
π2Γ 2
which coincides with the exact result.
56
CHAPTER 2. WKB APPROXIMATION
α=4
Since Γ
¡5¢
2
¡ ¢
­ 4®
2 1 Γ 52
x =√
(2n + 1)2 .
π 4 Γ (2)
=
3√
4 π,
­ 4® 3
x =
4
µ
¶
1
2
2n + 2n +
,
2
whereas the exact result seems to be (according to Landau)
­ 4®
¢
3¡ 2
x ex =
2n + 2n + 1 .
4
2.12
Exact WKB quantization
The WKB quantization condition can be actually converted into an algebraic
form, which can be used to calculate the eigenvalues to any order. The
method is due to Dunham and brought to its full glory by Bender, Olanssem
and Way.
In the following we will use their notation.
Starting from the Schrödinger equation written as
·
¸
d2
− 2 + V (x) − E ψ = 0,
ψ (±∞) = 0
dx
we introduce a small parameter ε and consider the eigenvalue problem
00
ε2 ψ = Q (x) ψ,
Q (x) = V (x) − E.
(2.11)
The parameter ε helps in organizing the WKB series and it os obviously
related to ~. The WKB solution is expressed as
" ∞
#
1X n
ψ (x) = exp
ε sn (x) .
ε
n=0
Substituting this expression into (2.11) and comparing the equal powers of
ε, we obtain
0
S (0) = − [a (x)]1/2 ,
0
0
2S0 Sn +
n−1
X
j=1
0
0
00
Sj Sn−j + Sn−1 = 0,
n ≥ 1.
(2.12)
2.12. EXACT WKB QUANTIZATION
57
The recursion equation in (2.12) is, as a matter of fact, a simple algebraic
0
rule for computing Sn0 (x) from sj for j < n. Straightforward computations
give
Q0 (x)
0
S1 (x) = −
,
4Q (x)
0
S2 (x) =
5 [Q0 (x)]2
00
−
Q (x)
5
3 ,
32 [Q (x)] 2
8 [Q (x)] 2
h 0
i3
0
00
000
15
Q
(x)
9Q (x) Q (x)
Q (x)
0
S3 (x) = −
+
−
,
64 [Q (x)]4
32 [Q (x)]3
16 [Q (x)]2
and so on. Once the Sn0 have been found, there is a simple formula which is
a generalization of the well–known formula of the WKB quantization
µ
¶
Z x2
1
1
π,
[E − V (x)] 2 dx = k +
2
x1
which states the exact quantization of the eigenvalues. This formula is
simply
I X
∞
1
Sn0 (x) = k
k = 0, 1, 2...
(2.13)
2πi
n=0
and it expresses, through the logarithmic derivative of the wave function,
the fact that this solution possesses k zeros. It is easy to see that the above
formula reduces to the usual one at the leading order. The leading WKB
0
formula takes into account both S0 (x) and S1 (x) substituting S0 (x) into
1
(2.13) and making [Q (x)] 2 single valued by joining the turning points x1
and x2 by a branch cut, we have
I
Z x2 p
1
1
−
[V (x) − E] 2 dx =
E − V (x)dx,
2πi
x1
0
for the term coming from S0 , whereas for the term coming from S1 (x)
I
1
Q0 (x)
1
1
1
−
dx = −
ln Q (z) |C = −
4πi = − ,
(2.14)
2πi
4Q (x)
8πi
8πi
2
Here ln Q (z) has been evaluated once around the contour (it gives 4πi because
the contour encircles two simple poles of Q (x)).
Then, one reproduces the well–known formula
Z x2 p
1
E − V (x) = k + .
2
x1
58
CHAPTER 2. WKB APPROXIMATION
An important comment is in order. It is remarkable that the above formula
(2.13) takes into account only the basic property of the wave function, i.e.
the number of nodes. Any reference, for instance, to the Airy function has
disappeared. Moreover, there is no reference to ψ (x) or to its boundary conditions. The construction is purely algebraic; it involves differentiation and
not integration. It is necessary though to use complex contour integration
instead of ordinary integration along the real axis between x1 and x2 . This
0
because all the functions Si (x) are singular at the turning points.
There is an interesting open question. Since the integrals in (2.13) involve
closed contours, it is possible to add total derivatives to Sn0 (z) under the
integral without altering the value of the integral. Hence, Sn0 as generated
by the recursion equation (2.12) is but one element of a large equivalence
class which we denote by Fn . The elements of Fn are all different but their
contour integral are always the same. Is it possible that for all n there is
some elements of Fn which is so simple that the indicate sum in (2.13) may be
evaluated in close form? This possibility seems quite remote. Nevertheless,
there are some interesting results.
0
(i) One element of F2n+1 is always zero (for n ≥ 1), because S2n+1 is
0
itself a total derivative. For example, S3 (x) can be written
 h

i2
0
00
d  5 Q (x)
Q (x) 
0
S3 (x) =
−

,
3
dx 64 [Q (x)]
16 [Q (x)]2
which vanishes once integrated along the close contour. This can be understood by a simple argument. The quantization condition (2.14) is a
0
constraint on the phase of ψ (x). S2n+1 (x) (n ≥ 1) is always real because
it contains no fractional powers of (V − E) and therefore cannot contribute
0 instead that becomes imaginary as x crosses
to the phase of ψ (x). It is S2n
into a classically allowed region and causes the wave function to become
0
oscillatory. Hence it is no longer surprising that S2n+1 (n ≥ 1) drops from
the quantization condition.
0 (z) , the only ones which survive, can be drasti(ii) The even terms S2n
cally simplified by adding and subtracting total derivatives. For instance
"
#
00
0
Q (x)
5Q (x)
d
0
S2 (x) = −
.
3 −
dx 48 [a (x)] 23
48 [a (x)] 2
(iii) There is a close parallelism between the search of the simplest element of F2n and the construction of the conserved quantities for the non–
2.12. EXACT WKB QUANTIZATION
59
linear wave equation like the KdV
·
000
0
u = u − 6uu .
For this equation, the conserved quantities are derived from the Miura transformation
0
u (x) = −v 2 (x) − v .
They satisfy a recursive equation, similar to (2.12).
The recursion equation for the elements Sn0 of the WKB expansion may
be derived from the Miura transformation if we put u = v/ε and let v =
P
0
εn Sn .
The Miura transformation if nothing but the Riccati equivalent of the
Schrödinger equation.
2.12.1
Some ”numerology”
Let us analyze the potential V = xN , with N even. For such potential, the
WKB series is a power series in inverse fractional powers of the energy E,
i.e.
µ
¶
∞
2
1
1 X
1
−n
1+
+
)
(
N
an (N ) = k +
π.
EN 2
E
2
n=0
The coefficients an (N ) have the form
¢
√ ¡
n
2 πΓ 1 + 1−2n
N ¢ Pn (N ) (−1)
an (N ) = ¡ 3−2n 1−2n
,
(2n + 2)!2n
Γ 2 + N
where Pn (N ) is a polynomial in N :
P0 (N ) = 1
P1 (N ) = 2 (N − 1)
P2 (N ) = (N − 3) (N − 1) (2N + 3)
P3 (n) =
¡
¢
4
(N − 1) (N − 5) 24N 3 + 22N 2 − 117N − 139
9
..
.
All these polynomials contain (N − 1) as a factor, and this explains why for
the harmonic oscillator the WKB quantization is exact.
60
CHAPTER 2. WKB APPROXIMATION
One should keep in mind that the above series is an asymptotic series.
Like the Stirling series for the Γ function, the coefficients get smaller for a
while but eventually grow without bound. This means that for any given
value of k, successive approximations to E (k) , obtained by truncating the
series, improve to some maximal accuracy and then become worse. Also,
since E increases with k, more terms in the series should be required to reach
maximal accuracy as k increases, and the accuracy should also increase with
k. This is precisely what happens. For instance, with V = x4 ,
0
Eexact
= 1.060362090
(W KB)1 = 0.87
(W KB)2 = 0.98
(W KB)4 = 0.95
(W KB)6 = 0.78
(W KB)8 = 1.13
(W KB)10 = 1.40,
8
Eexact
= 37.923001027033
(W KB)1 = 37.904471845068
(W KB)2 = 37.923021140528
..
.
(W KB)10 = 37.923001027043
2.13
¡ −14 ¢
10 ! .
Estimate of the ground state energy
In the following we will discuss some useful tools which allows us to estimate the ground state energy, i.e. to have upper and lower bounds for this
quantity.
Variational principle
The Hamiltonian can be decomposed as
X
H=
En |ψn i hψn | .
(2.15)
n
2.13. ESTIMATE OF THE GROUND STATE ENERGY
61
In the following we assume to have an increasing sequence of eigenvalues
and, for simplicity, to have no degeneracy. Taking the expectation value of
(2.15) on a state |ψi, we have
X
X
hψ| H |ψi =
En |hψ| ψn i|2 ≥ E0
En |hψn | ψi|2 = E0 hψ| ψi , (2.16)
n
n
i.e.
E0 ≤
hψ| H |ψi
.
hψ| ψi
(2.17)
If the state depends on a parameter, i.e. |ψi = |ψ (β)i and it has the same
qualitative features of the ground state wave function (i.e. it has no nodes
and decreases sufficiently rapidly at infinity), we can obtain an estimate of
E0 by computing (2.17) and determining the value minimizing this quantity.
Example
· 2
¸
p2
pb
a
a
H=
+ λ |x| = ε0
+ b |η| ,
(2.18)
2m
2
where we have chosen the natural units, previously introduced, which match
the semiclassical expression. Let us take now, as a trial wave function,
s
µ
¶
β
β 2η2
ψβ (η) = √ exp −
.
2
π
By using the formula
Z
I (n) =
+∞
|x|n e−α
2 x2
dx =
−∞
¶
µ ¶n+1 µ
1
n+1
,
Γ
α
2
it is easy to obtain the expectation value of the Hamiltonian. In fact, we
have,
0
00
ψ = −β 2 ηψ;
ψ = β 4 η 2 ψ − β 2 ψ.
Therefore, for the kinetic term we get
E β2 β4
p2
1
00
hψ| |ψi = − hψ| ψ =
−
hψ| η 2 |ψi =
2
2
2
2
β2 β2
β2
−
=
,
2
4
4
whereas for the potential term we have
¢
¡
b Γ a+1
B
2
√
hψ| V |ψi = b hψ| |η| |ψi =
= ,
β
β
π
a
62
CHAPTER 2. WKB APPROXIMATION
where
b (a)
B= √ Γ
π
Hence
µ
·
hψ| H |ψi = ε0
a+1
2
¶
.
¸
β2
B
+ a .
4
β
(2.19)
Minimizing with respect to β, we have
¸
·
∂
β
B
hψ| H |ψi = ε0
− a a+1 = 0,
∂β
2
β
i.e.
1
βop = (2aB) a+2 .
Substituting now into (2.19), we have
"
#
µ
¶
2
2aB
1
B
1
a+2
= ε0
=
E0 ≤ ε0
(2aB)
+
+B
1
1
4
4
(2aB) a+2
(2aB) a+2
µ
ε0
a+2
4a
¶
(2aB)
so, finally
2
a+2
µ
E0 ≤ ε0
µ
= ε0
a+2
4a
¶·
a+2
4a
¶·
2ab (a)
√ Γ
π
2ab (a)
√ Γ
π
µ
a+1
2
µ
¶¸
a+1
2
2
a+2
¶¸
2
a+2
,
.
For a = 2, we get correctly E0 = ε20 .
The variational principle can be used to establish the following general result:
an attractive potential, i.e. a potential V (x) < 0, has always at least a bound
state.
2.14
Uncertainty relation and lower bound
The previous variational principle permits to obtain an upper bound for
the ground state energy E0 . In order to obtain a lower bound, we rely on
Heisenberg’s uncertainty relation
∆p∆q ≥
~
.
2
For an Hamiltonian as the one in (2.18), we have
" µ
#
¶
1
1 2
a
E0 ≥ ε0
+ b |η| .
8 ∆q
2.14. UNCERTAINTY RELATION AND LOWER BOUND
63
Requiring that ∆q is of the order of 2η we have
" µ
¶
µ
¶ #
1
∆q a
2 2
E0 ≥ ε0
+b
≡
32 ∆q
2
"
ε0
where x ≡
have
∆q
2 .
1
32
#
µ ¶2
1
+ bxa ,
x
(2.20)
Computing the extremum of the above quantity w.r.t. x we
1
−
16
µ ¶3
1
+ abxa−1 = 0,
x
µ
⇒ xop =
1
16ab
¶
1
a+2
.
Substituting this value into (2.20), we have, after some manipulations
µ
¶
2
a+2
E0 ≥ ε0
(16ab) a+2 .
32a
For the ratio of the uncertainty bound and the one of the variational principle
we have
à √ ! 2
a+2
Eun
1
8 π
¡ a+1 ¢
=
,
Evar
8 Γ 2
and this quantity is correctly always less than 1.
64
CHAPTER 2. WKB APPROXIMATION
Chapter 3
Thermodynamic Bethe
Ansatz and Ordinary
Differential Equations
3.1
Spectral determinant
In this section we will come back to the problem of the spectral determinant of the Schrödinger equation. We will see that for a particular class of
potentials this function satisfies a nonlinear functional equation which, as a
matter of fact, is equal to the Bethe Ansatz equations for integrable models
in Statistical Mechanics! Consider the Schrödinger equation
µ
¶
d2
− 2 + P (x) ψ (x) = 0
(3.1)
dx
for arbitrary complex values of x, with
l (l − 1)
.
x2
For non integer values of M we have a branch cut at the origin. Using the
results of the analysis by Sibuya in [?], we state that (3.1) admits a solution
y = y (x, E, l), such that
(i) y is an entire function of x, E, where due to the branch point at x = 0,
x must be in general considered to live on a suitable cover of the punctured
complex plane;
0
(ii) y and y admit the asymptotic expressions
·
¸
1
−M
M
+1
y ∼ x 2 exp −
x
(3.2)
M +1
P (x) = x2M − E +
65
66
CHAPTER 3. TBA AND DIFFERENTIAL EQUATIONS
·
M
y ∼ −x 2 exp −
¸
1
xM +1 ,
M +1
(3.3)
as x goes to infinity in any closed sector satisfying
|arg x| <
3π
.
2M + 2
(Remark. Extra terms appear though for 0 < M ≤ 1 ).
(iii) Furthermore, the solution y is uniquely identified by the above information. From WKB theory, we know that eq. (3.1) has two solutions Φ±
given by
· Z x
¸
p
− 41
Φ± (x) = [P (x)] exp ±
P (x)dx .
x0
It is then easy to see that the asymptotic (ii) is, up to a normalization
constant N , the large x limit of Φ− :
y (x, E) ' N Φ− (x, E)
with x = ρeiθ we have
− π2
Φ± (ρ, θ) ' x
¸
·
ρM + 1 iθ(M +1)
e
.
exp ±
M +1
Let us fix now some terminology. A solution tending to zero for large ρ in
the sector
a<θ<b
is called subdominant in that sector. A solution growing for large ρ in the
sector
0
0
0
a <θ <b
is called dominant in this sector.
Let us denote by Sk the sector
¯
¯
¯
¯
π
¯θ − 2kπ ¯ <
.
¯
¯
2M + 2
2M + 2
We see then
(i) in S0 , y ' N Φ− → 0, i.e. y is subdominant in S0 ;
(ii) at θ = ± 2Mπ+2 it decays algebraically;
(iii) in S±1 , y ' N Φ− → ∞, i.e. y is dominant in these sectors.
FIGURE
3.1. SPECTRAL DETERMINANT
67
Note that in S±1 , with some coefficients C (E)
y (x) = y− (x) + C (E) y+ (x)
and that exactly at θ ≥ 2M3π+2 we have lost control of the exponential asymptotic behaviour of y. This is the reason of the above restriction
|θ| = |arg x| <
3π
.
2M + 2
To find subdominant solutions on other sectors, we can use the following
technique. Consider yb (x) = y (ax, E, l) for any constant a. This function
satisfies the differential eq.
¸
·
l (l + 1)
d2
− 2 + a2M +2 x2M − a2 E +
yb (x) = 0.
dx
x2
¡ ¡
¢¢
Thus, if a2M +2 = 1, P y ax, a−2 E, l is another solution of (3.1). Setting
ω=e
iπ /M +1
,
we have therefore the set of solutions
³
´
k
yk ≡ yk (x, E, l) = ω 2 y ω −k x, ω 2k E, l
(3.4)
k
with yk subdominant in Sk and dominant in Sk±1 (the prefactor ω 2 is for
later convenience). The first conclusion of the above analysis is that each
pair
{yk , yk+1 }
of functions provides a set of linearly independent solutions of (3.1) and any
other solution can be expressed in terms of them. In particular
ek (E, l) yk+1 (x, E, l) .
yk−1 (x, E, l) = Ck (E, l) yk (x, E, l) + C
(3.5)
fk are called the Stokes multipliers for yk−1 w.r.t.
The functions Ck and C
yk and yk+1 . Clearly, from (3.4), we have
¡
¢
Ck (E, l) = Ck−1 ω 2 E, l
¡ 2
¢
fk (E, l) = C
]
C
k−1 ω E, l .
f0 by C and C.
e The second result is that
From now on, we denote C0 and C
the Stokes multipliers can be expressed in terms of the Wronskians.
68
CHAPTER 3. TBA AND DIFFERENTIAL EQUATIONS
Definition 1. The Wronskian W [f, g] of two functions is defined by
0
0
W (f, g) = f g − f g.
If f and g are both solutions of the Schrödinger equation (3.1), their Wronskian is a constant, independent of x. In fact
0
0
00
00
0
0
00
00
W 0 (f, g) = f g +f g −f g −f g = f g −f g = −P (x, E, l) (f g − gf ) = 0.
Consider the Wronskian of (3.5) (with k = 0) with respect to the functions
y1 and y0 :
f0 y1
y−1 = C0 y0 + C
W [y−1 , y1 ] ≡ W−1,1 = C0 W [y0 , y1 ] = C0 W01
h
i
f0 W y1 , y0 = C
f0 W1,0 = −C
f0 W01 ,
W [y−1 , y0 ] ≡ W−1,0 = C
i.e.
C=
W−1,1
;
W01
e = − W−1,0 .
C
W0,1
(3.6)
These Wronskians are entire functions of E and l (since the same holds
for the functions yk ). Since y0 and y1 are linearly independent, W never
e are also entire functions. As a matter of fact,
vanishes. Therefore C and C
f
all Ck are identically equal to −1,
fk = −1.
C
This result follows from the relations
¡
¢
Wk1 +1,k2 +1 = Wk1 k2 ω 2 E, l
W0,1 (E, l) = 2i.
The last condition is obtained by evaluating W0,1 as x → ∞ in the sectors S0
or S1 , where the asymptotic behaviours of y0 and y1 (and their derivatives)
are determined by eqs. (3.2)–(3.3). Since y−1 (x, E, l) = y1 (x∗ , E ∗ , l∗ ) it
also follows from (3.6) that C (E, l) is real whenever E and l are real.
Finally, the basic Stokes relation (3.5) at k = 0 can be written in the form
C (E, l) y0 (x, E, l) = y−1 (x, E, l) + y1 (x, E, l)
C (E, l) =
1
W−1,1 (E, l) ,
2i
(3.7)
3.1. SPECTRAL DETERMINANT
69
i.e.
¡
¢
¡
¢
1
1
C (E, l) y (x, E, l) = ω − 2 y ωx, ω −2 E, l + ω 2 y ω −1 x, ω 2 E, l .
Let us analyze this equation. Any solution of (3.1) can be written in terms
of another basis of functions
{ψ+ , ψ− }
where ψ+ (x) is the solution which for x → 0 goes as
ψ+ ∼ xl+1
whereas ψ− (x) behaves as
(3.8)
ψ− ∼ x−l .
Notice that both of them vanish if l (l + 1) < 0, whereas one of the two
diverges if l (l + 1) > 0.
Since l enters eq. (3.1) only in the combination l (l + 1), we can decide to
fix ψ+ (x) as the solution which behaves as (3.8) and to define ψ− (x) by its
analytic continuation, i.e.
ψ− (x, E, l) ≡ ψ+ (x, E, −1 − l) .
In analogy with what done previously, we define the shifted solutions ψk±
³
´
k
ψk± = ψk± (x, E, l) = ω 2 ψ ± ω −k x, ω 2k E, l .
They also solve the original problem (3.1). By considering the limit x → 0,
it is easily seen that
ψk± (x, E, l) = ω ∓k(l+ 2 ) ψ ± (x, E, l) .
1
We also have
h
i
h
i¡
¢
W yk+1 , ψk±2 +1 (E, l) = W yk+1 , ψk±2 +1 ω 2 E, l
and therefore
£
¤
£
¤
1
W yk , ψ ± (E, l) = ω ±k(l+ 2 ) W yk , ψ ± (E, l) =
´
£
¤³
1
ω ±k(l+ 2 ) W y, ψ ± ω 2k E, l .
(3.9)
We can now take the Wronskian of both sides of (3.7) with ψ ± . Defining
£
¤
D∓ (E, l) ≡ W y (x, E, l) , ψ ± (x, E, l)
70
CHAPTER 3. TBA AND DIFFERENTIAL EQUATIONS
and using (3.9), the Stokes eq. (3.7) becomes
¡
¢
¡
¢
1
1
C (E, l) D∓ (E, l) = ω ∓(l+ 2 ) D∓ ω −2 E, l + ω ±(l+ 2 ) DD∓ ω 2 E, l . (3.10)
We will discover soon the meaning and the beauty of this functional equation.
First of all, notice that expressing
y = αψ + + βψ − ,
the coefficients of this linear combination are given by the above Wronskians.
In fact, using
£
¤
W ψ − , ψ + = x−l (l + 1) xl + lx−l−1 xl+1 = 2l + 1,
we have
£
¤
W y, ψ + ≡ D− = β (2l + 1)
£
¤
W y, ψ − ≡ D+ = −α (2l + 1) ,
i.e.
(2l + 1) y = D− (E, l) ψ − − D+ (E, l) ψ + .
Moreover, from the relation
ψ − (x, E, l) = ψ + (x, E, −1 − l)
we deduce
D (E, l) ≡ D− (E, l) = D+ (E, −1 − l) .
Requiring a non–singular behaviour at the origin for y (x) restricts E to the
set of eigenvalues {En } of the spectral problem, i.e.
D (E, l) |E=En = 0,
1
l>− .
2
Therefore we arrive at the following remarkable conclusion: D (E, l) is an
entire function which has zeros on the real positive axis, in correspondence
with the eigenvalues EM of the Schrödinger equation. This because the
solution y vanishes at infinity (x → ∞) and it is regular as x → 0 . Hence
D± (E, l) are the Fredholm determinants. We have the following properties.
(i) C and D− are entire functions of E, since the functions entering
their definition are entire.
(ii) If l is real and larger than − 12 , then the zeros of D− all lie on the positive real axis of the complex plane E. In fact, we have already commented
3.1. SPECTRAL DETERMINANT
71
that a zero of D− (E, l) signals the existence of a proper eigenfunction. The
Hermitian nature of the problem for l > − 12 then ensures the reality of these
zeros.. Moreover, for l > 0 the potential is everywhere positive. Therefore,
by multiplying for ψ ∗ and integrating from (0, ∞) shows that all eigenvalues
EM must be positive.
(iii) When M > 1, D− (E, l) has the following large E asymptotic
ln D− (E, l) '
a0
(−E)µ ,
2
|E| → ∞,
|arg (−E)| < π,
(3.11)
where
µ
¶ µ
¶
1
1
1
1
a0 = − √ Γ − −
Γ 1+
,
2 2M
2M
π
µ=
M +1
.
2M
(iv) If E = 0, then
¶
µ
2l+1
1
1
2l + 1
(2M + 2) 2M +2 + 2 .
D (0, l) = − √ Γ 1 +
2M + 2
π
−
Notice that at l = 0, we simply have
D− (E, 0) ≡ y (x, E) |x=0 , D+ (E, 0) ≡ y 0 (x, E) |x=0 .
In this case, the problem reduces to an ordinary Schrödinger equation, but
on a half line. WE can choose Dirichlet b.c. for the solution, i.e.
y (x, E) |x=0 = 0,
or Neumann b.c.
0
y (x, E) |x=0 = 0.
The first one are those appropriate for the wave function in the odd–parity
sector, as for instance
FIGURE
whereas the second one refers to the even–parity sector, as
FIGURE
In this case D− (E, 0) will have zeroes in correspondence of the energy levels
En− with odd parity, whereas D+ (E, 0) will have zeros in correspondence of
En+ . This is equivalent, in fact, to choose the proper combination which
expresses y in this case.
72
CHAPTER 3. TBA AND DIFFERENTIAL EQUATIONS
For l = 0, it is simple to establish the formula (3.11). In this case, in fact,
D− (E, 0) ≡ y (x, E) |x=0 = 0, so we have to find the value of this function
at the origin. When E → ∞, we have
P 0 (x, E)
→ 0,
P (x, E)
and therefore we can rely on WKB. We have, in fact
y (x, E) ' N φ− (x, E)
and therefore
y ∼
ln
=
N
Z
x0
¡ 2M
¢1
E
− E 2 dE
x
'
x,x0 À0
+1
xM
xM +1
0
−
.
M +1 M +1
We can drop the first term by choosing
Z x0
ln N = −
tM dt.
0
Sending now x0 → ∞, we end up with
¸
Z ∞·
¡ 2M
¢ 12
M
ln y | x=0 ∼
t −E −t
dt = a0 (−E)M ,
.EÀ0
0
where
Z
a0 =
0
∞ µ£
¤1
t2M + 1 2 − tM
¶
µ
¶ µ
¶
1
M +1
1
dt = − √ Γ −
Γ 1+
,
2M
2M
π
with
M +1
.
2M
The same result can be shown to hold also for l 6= 0. We can now use the
Hadamard factorization theorem.
µ=
3.2
Hadamard Factorization Theorem
An entire function f (z) is said to be of finite order if there is a positive
number A such that
³ A´
|f (z)| = O e|z| ,
|z| → ∞.
3.2. HADAMARD FACTORIZATION THEOREM
73
The lower bound δ of the numbers A for which this is true is called the order
of f (z). In our specific example,
ln D (E, l) |−EÀ0 ∼ a0 (−E)M
which is valid everywhere except for arg (E) = 0, i.e. the real axis, where
the growth of D is not greater than for arg (E) 6= 0. Hence
order [D] = µ =
M +1
< 1,
2M
M > 1.
The celebrated Hadamard theorem can be stated as follows.
Theorem 1. An entire function f(z) of finite order δ admits a factorization
over his zeros {zn } of the form
f (z) = e
Q(z)
∞
Y
µ
K
n=0
z
,P
zn
¶
,
where
2
K (v, 0) = 1 − v;
p
v+ v2 +...+ vp
K (v, p) = (1 − v)
are called “primary factors” and Q (z) is a polynomial of order not greater
than δ. If δ is not an integer, then p = Int [δ]. Consequently, for M > 1,
δ=µ=
and
M +1
,
2M
p=0
¶
∞ µ
Y
E
.
D (E, l) = D (0, l)
1−
En
n=0
For M ≤ 1 things are complicated by the possibility of nontrivial Q (z).
Exactly at M = 1,
¶
∞ µ
Y
E
E
D (E, l) = D (0, l)
1−
e En .
En
n=0
74
CHAPTER 3. TBA AND DIFFERENTIAL EQUATIONS
3.3
Bethe Ansatz Equations
The functional eq. (3.10) can be used to derive the energy levels of the
considered quantum mechanical problem. Let us consider the one for D− ≡
D
¡
¢
¡
¢
1
1
C (E, l) D (E, l) = ω −(l+ 2 ) D ω −2 E, l + ω (l+ 2 ) D ω 2 E, l .
Setting E = Ei the left–hand side vanishes and we are led to the equation
¡ 2
¢
2l+1 D ω Ei , l
−1 = ω
;
ω ≡ eiπ/M +1 .
D (ω −2 Ei , l)
For M > 1, we can use the Hadamard factorization theorem and write
−1 = ω
2l+1
¶
∞ µ
Y
En − Ei ω 2
.
En − Ei ω −2
n=0
This is an equation for the energy levels, equivalent to the Bethe Ansatz
equation in Statistical Mechanics! This equation can be converted into an
integral nonlinear equation, which can be solved very fast numerically. To
this aim, let us define
¡
¢
¶
∞ µ
Y
D Eω 2
En − Eω 2
−2l−1
a (E, l) ω
=
=
.
(3.12)
D (Eω −2 )
En − Eω −2
n=0
When a (E, l) = −1, we get the quantization condition for the energy levels.
From eq. (3.12), taking the logarithm we have
∞
X
2l + 1
π+
F
ln Q (E, l) = i
M +1
n=0
with
µ
E
En
¶
,
1 − ω2E
.
1 − ω −2 E
Assuming that all En lie on the positive real axis, and that these are the only
points in some strip about this axis for which a (E, l) = −1. By Cauchy’s
theorem we have
µ ¶
Z
¡
¡
¢¢
2l + 1
dE 0
E
ln a (E, l) = i
π+
F
∂E 0 ln 1 + a E 0 , l .
0
M +1
E
C 2πi
F (E) = ln
FIGURE
3.3. BETHE ANSATZ EQUATIONS
75
since
¡
¡
¢¢
∂E 0 ln 1 + a E 0 , l =
a0 (E 0 )
1 + a (E 0 )
and then apply the residue theorem. We make the change of variables
θ
E = eµ
and integrate by parts. We get
Z
¡
¢ ¡
¡ ¢¢
2l + 1
π+
dθ0 R θ − θ0 ln 1 + a θ0 +
M +1
C
Z
¡
¢ ¡
¡ ¢¢
dθ0 R θ − θ0 ln 1 + a θ0 ,
−
ln a (θ) = i
C
where the contour C1 and C2 run from −∞ to +∞, infinitesimally below
and above the real axis, and
R (θ) ≡
i d ³ µθ ´
F e .
2π dθ
FIGURE
Using the fact that i ln a is a real analytic function for which
[a (θ)]∗ = [a (θ∗ )]−1 ,
and pushing C1 and C2 toward the real axis, we have
Z +∞
¡
¢n h
¡
¢−1 i∗
¡
¡
¢¢o
2l + 1
π+
dθ0 R θ − θ0 ln 1 + a θ0 − iε
ln a (θ) = i
− ln 1 + a θ0 − iε
M +1
−∞
2l + 1
π+
=i
M +1
Z
+∞
¡
¢£
¡ ¢
¡
¡
¢¢¤
dθ0 R θ − θ0 ln a θ0 − 2iIm ln 1 + a θ0 − iε .
−∞
Let us define the Fourier transform
Z
fe(k) = F (f ) (k) =
+∞
f (θ) e−ikθ dθ
−∞
f (θ) = F
−1
Z +∞
h i
1
e
fe(k) eiθk dk,
f (θ) =
2π −∞
76
CHAPTER 3. TBA AND DIFFERENTIAL EQUATIONS
and taking the Fourier transform of the above equation
h
i
e (k) F [ln a] (k) = 2iπ (2l + 1) δ (k) − 2iR
e (k) ImF [ln (1 + a)] (k) .
1−R
M +1
Transforming back and restoring the integrations over C1 and C2 , we have
µ
¶
1
ln a (θ) = iπ l +
− ib0 eθ +
2
Z
Z
¡
¢ ¡
¡ 0 ¢¢
¡
¢ ¡
¡ ¢¢
0
0
0
+
dθ ϕ θ − θ ln 1 + a θ −
dθ ϕ θ − θ0 ln 1 + a−1 θ0 ,
C1
C2
with
Z
dk ikθ sh π2 (ξ − 1) k
1
e
,
ξ=
.
π
πk
M
2sh 2 ξkch 2
−∞ 2π
π
b0 = 2 cos
a0 .
2M
The driving force ib0 eθ of the above equation arises from a zero mode, which
can be traced to the zero of
+∞
ϕ (θ) ≡
e (k)
1−R
at
K = i.
Its value can be fixed by the WKB result
D (E, l)
'
E→−∞
³ θ ´n
a0 (−E)n = a0 −e µ
and from the definition
a (E, l) = ω
2l+1
¡
¢
D Eω 2 , l
.
D (Eω −2 , l)
A first consistency check of this equation is immediate in the large θ limit,
the driving term dominates and therefore
a (θ, l) = −1
at the points θn , with
b0 e
giving
θn
¶
µ
1
+ (2n + 1) π,
'π l+
2
·
n = 0, 1, . . .
¸1
n
π
En '
(4n + 2l + 3) ,
n = 0, 1, . . .
2b0
which at M = 1, a potential can be solved exactly, coincides with the exact
result!
Part II
Analytic Number Theory: an
Introduction
77
Chapter 4
Prime numbers
4.1
Introduction
Number theory is one of the most fascinating areas of Mathematics. There
are many unproved results and conjectures. The most amazing aspect is
that there are problems which can be stated in a very simple way, but very
difficult to prove.
Example: (3n + 1) Problem.
Take a number N. If it is even divide it by 2, otherwise multiply it by 3 and
add 1.
½
N→
¯
N/2,
N even ¯¯
.
3N + 1,
N odd ¯
Iterate the operation. Does the numbers always end to 1?
(Argument based on random walk).
Number theory has deep overlaps with calculus and also with geometry.
Example: Pitagoric triple.
a2 + b2 = c2 ,
a, b, c ∈ N.
a = 3, b = 4, c = 5,
a = 5, b = 12, c = 13,
a = 8, b = 15, c = 17,
..
..
..
.
.
.
79
80
CHAPTER 4. PRIME NUMBERS
There is an infinite number of them. Geometrically, the problem consists in
finding all rational points on a circle
³ a ´2
c
µ ¶2
b
+
= 1.
c
FIGURE
This happens when the slope is rational! The generalization of this problem
consists in finding rational points on an elliptic curve. According to the
recent proof of Fermat’s last theorem, there are no integer solutions to the
equation
an + bn = cn ,
n > 2.
Another classical connection between Number Theory and Geometry is
about the possibility to construct regular polygons by rule and compass.
According to the proof by Gauss, this is possible for all those polygons of n
sides when n is a product of different Fermat primes and powers of 2:
n
Fn = 22 + 1.
At present, only 5 such primes are known for n = 0, 1, 2, . . ., i.e.
F0 = 3,
F1 = 5,
F2 = 17,
F3 = 257,
F4 = 65537.
So, regular polygons with
n = 2, 3, 4, 5, 6, 8, 10, 12, 15, 17, 20, 24, 30, . . .
sides are constructible while those with
n = 7, 9, 11, 13, 14, 18, 19, 21, 22, 23, 25, . . .
cannot be constructed.
In these lectures, we will cover only a small set of topics of this large subject.
In the first lecture we will discuss some beautiful identities, collected and
presented almost in a random order. The other two lectures will deal with
one of the most fascinating and mysterious functions of Mathematics, i.e.
the Riemann zeta function. We will discuss its main properties and its
relationship with prime numbers.
4.2. INDUCTION
4.2
81
Induction
One of the strongest properties of integers is their inductive nature: if the
assumption that some law holds for n implies that holds for n + 1 as well
(and it is proved true for n = 1), then it is always true.
A typical result that can be easily proven by induction is
n
X
n (n + 1)
.
2
k=
k=1
Another, less well known, is the derivation by Newton of the infinite expansion of the series
(1 + x)−2 = 1 − 2x + 3x2 − 4x3 + 5x4 + ...
(1 + x)−3 = 1 − 3x + 6x2 − 10x3 + 15x4 + ...
Consider the Tartaglia triangle for the binomial expression (1 + x)n , with n
nonnegative integer:
n=0
1
n=1
1
n=2
1
n=3
n=4
1
2
1 3
1
1
3
4 6
1
4
1
Of course, each number is the sum of the two numbers above it, i.e.
µ
¶ µ
¶ µ
¶
n+1
n
n
=
+
.
k
k−1
k
By applying this rule back–forward, we get
n = −3
1
−3
6
− 10 15
−4 5 −6
n = −2
1
−2
3
n = −1
1
−1
1 −1
n=0
1
0
0
1−1
0
Let us consider now some limiting formulas
r
√
q
√
1+ 5
1 + 1 + 1 + ... →
2
0
82
CHAPTER 4. PRIME NUMBERS
r
q
7+
s
√
1+
7 + 7 + ... →
√
29
2
r
q
√
1 + 2 1 + 3 1 + 4 ... → 3
s
q
1+
2+
r
q
s
1+
4.3
r
2+
3+
√
√
4+... → 3
√
... pn +... → 1.8507...
Prime Numbers
Let us consider now the Prime Numbers pi , the building blocks of Number
Theory. The first primes are
pi = 1, 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, ...
There are infinitely many prime numbers. Several proofs of this fact are
known; the oldest and simplest one is due to Euclid.
Euclid’s proof. Assume that there are only k primes p1 , ..., pk . Consider
the number N = p1 · ... · pk + 1. It is not divisible for any pi , so either it is
a prime or it has a prime divisor larger than all p1 , ..., pk .
The basic theorem of Arithmetic states that any natural number n either is
a prime or admits an unique decomposition in terms of the primes, i.e.:
n = pα1 1 · ... · pαk k .
The mathematical problem of identifying a prime number or to determine
its prime decomposition is one of the most challenging of Mathematics.
Let us discuss briefly the Primality Problem. The crucial point is that it
doesn’t exist a formula which determines all and only the primes, i.e. a
relation of the type Pn = f (n). There are several formulas providing an
answer to the primality problem, but in practice almost useless.
Example. Wilson formula
·
¸
(n − 1)! + 1
f (n) = sin π
.
n
One can prove that this function is zero if and only if n is a prime. Unfortunately, this primality test is of no practical advantage, because calculating
f (n) takes longer than the ordinary Eratosthenes’s sieve.
Example: 101! ' 10160 .
4.4. MEASURE OF COMPLEXITY
4.4
83
Measure of Complexity
Consider an algorithm which applies to a problem made of N data. The
complexity of the problem can be related to the time of elaboration, in
particular how this time scales with N . The problem has a polynomial
complexity if the time of elaboration scales as a power law of N :
τpoly ∼ N α .
These are the “easy” problems of mathematics. On the other hand, the
problem is said to have an exponential complexity if the time scales faster
than any power of N
τexp ∼ exp [βN γ ] .
These are the “toughest” problems of mathematics. Among them:
• the problem of salesman,
• the determination of the vacuum states in random systems.
Consider now a number A, expressed, say, in the usual decimal basis. The
number of its digits is about log10 A. In this context, a polynomial algorithm
scales as
τpoly ∼ (log10 A)α ,
whereas an exponential algorithm scales as the number A itself, since
τexp ∼ exp [β log A] ∼ Aβ .
Let us discuss then the primality test. In its simplest version it consists in
the Eratosthenes’s sieve. This means,
√ in practice, testing the divisibility of
the number A for all numbers B ≤ A, hence
τ ' A1/2 ,
and we have an exponential algorithm! Only recently, it has been proved by
Agrawal et al. that there exists a primality test with polynomial complexity.
The same story happens for the factorization problem. In fact, is its naive
implementation, the factorization algorithm scales as well as τ ' A1/2 . Until
now, it doesn’t still exist a factorization algorithm with polynomial complexity.
Let us see how Quantum Mechanics may help in this respect.
84
4.5
CHAPTER 4. PRIME NUMBERS
A Primality Test in Quantum Mechanics
Suppose we have constructed a potential V (x) which has as eigenvalues the
prime numbers sequence (this can be reasonably done in a semi–classical
approximation). We can construct an algorithm for a primality test with
zero complexity, i.e. it does not scale or depend on the number of digits of
the number A under check!
FIGURE
Let us denote by G (in honour of Gauss) the device furnishing the primality
test. If a wave of energy E = ~ωA is transmitted, A is a prime. Otherwise,
it is composite, and it remains the problem of determining its factors :
A = p1 · ... · pk .
For simplicity, assume they are all different; the following discussion can be
easily generalized. The question is: how Quantum Mechanics does solve the
factorization problem?
4.6
Factorization Problem in Quantum Mechanics
Quantum Mechanics works well with sums of numbers rather than with
products. Hence take initially the logarithm of the numbers under check
and promote them to energies. From the unique factorization theorem, we
know that any number can be written in an unique way as a product of
primes: A = p1 · ... · pk , so
log A =
k
X
log pi .
i=1
In the last expression, single out one term (any of them), and write
log A = log pbj +
k
X
bj ,
log pi ≡ log pbj + log N
i6=j
bj = A/pj .
where N
Let us construct now (by the inversion formula) a potential Vlog p (x) which
possesses as eigenvalues all and only the logarithms of primes. Let us construct also another potential Wln N (y) whose eigenvalues, instead, are all
4.6. FACTORIZATION PROBLEM IN QUANTUM MECHANICS
85
the logarithms of the integer numbers. Consider now the 2–dimensional
Hamiltonian
H (x, y) =
¤
1 £ 2
px + p2y + V (x) + W (y) .
2m
With respect to this Hamiltonian, the eigenvalues associated with a number A of k prime factors is k–times degenerate. This is obvious since one
can choose any term ln pbj in the previous decomposition. Denote the k–th
element of the basis of this degenerate level as
¯ E
¯b
|b
pj i ¯N
j .
Suppose that we switch on an energy precisely equally to log A into the
system. According to Von Neumann’s theorem, the system immediately
after will be in a linear combination of these k states:
|ψ (t)i =
k
X
cj |pj , Nj ; ti .
j=1
This state, under time evolution, remains always in the original degenerate
subspace.
Suppose now that we send on the system an electromagnetic wave, polarized
precisely along the y–axis. In a dipole approximation the system couples to
the operator yb. This operator induces a transition among the levels of the
W (y) potential, in particular the most important ones are between the next
neighborhoods
|ln N + 1i
%
|ln N i
&
|ln N + 1i
and the spectrum of emission/absorption of the system becomes the one of
k multiplets.
FIGURE
If the original system was an harmonic oscillator, the splitting of these multiplets would be always the same, since the energy levels of the harmonic
oscillator are equally spaced.
FIGURE
86
CHAPTER 4. PRIME NUMBERS
But is this case the potential along y has eigenvalues which are log N , hence
∆E± = ln
(N ± 1)
1
'± ,
N
N
i.e. the separation of the energy levels contains information on the level. In
this way, we have two basic informations.
1. How many factors the number A is made of, since it is equal to the
number of multiplets,
bi , taken the largest one (i.e. the
2. From their splittings we can derive 1/N
b
bi . Dividing the
one associated to the smallest Ni ), one determines N
bi , one extracts pbi and the problem has reduced
original number by N
its difficulty to a step less.
Continuing in this way, in k measurements one ends up with a complete
factorization of the original number A.
4.7
Elementary facts about prime numbers
• Each odd prime number is always of the form
4n ± 1.
• All prime numbers of the form 4n + 1 can be written as sum of two
squares, while the other ones cannot.
4.8
Fermat’s little theorem
This is a famous result in elementary number theory.
Theorem 2. If a is any integer and p is a prime, then ap − a is divisible
by p, or
ap−1 ≡ 1
(modp) .
This theorem has been generalized by Euler. He introduced the function
φ (m), called Euler’s φ function or totient function. It counts the number
of positive integers r smaller than m that are coprime to m.
Example. m = 10, r = 1, 3, 7, 9, φ (10) = 4.
4.8. FERMAT’S LITTLE THEOREM
87
For a prime number p, each of the previous r = 1, 2, ..., p − 1 is coprime to
p, hence
φ (p) = p − 1.
It is easy to prove that
α
φ (p ) = (p − 1) p
α−1
¶
µ
1
=p 1−
.
p
α
In fact, among the pα numbers less than the number φ (pα ), pα−1 are divisible
by p, hence the result.
The Euler function is a multiplicative function, i.e.
φ (n · m) = φ (n) φ (m)
if (n, m) = 1.
Q
Therefore, it is easy to calculate for any number m = i pεi i
¶
Y
Y
Yµ
1
εi −1
εi
(pi − 1) pi
=m
1−
φ (m) =
φ (pi ) =
.
pi
i
i
i
Notice that this function depends on the multiplicative properties of the
considered number so, it has wild jumps
φ (n) = 1, 1, 2, 2, 4, 2, 6, 4, 6, 4, 10, 4, 12, 6, 8, 8, 16, ...
n=
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17...
With the above definitions, the Euler theorem states that
bφ(m) ≡ 1
(modm)
if
(b, m) = 1.
We will estimate later on the behaviour of φ (m) and its average properties.
Note that
µ
¶
1
φ (n) Y
1−
=
.
n
pi
pi |n
We can convert this product over primes dividing n into a product over all
primes by a probabilistic argument. Indeed, the probability that any “old”
prime will divide n is 1/pi and the probability that it will not is 1 − 1/pi .
In the latter case, the prime pi “contributes” the factor 1 to the product.
Hence
¶ Y ·µ
¶
µ
¶¸ Y µ
¶
Yµ
1 1
1
1
1
'
1−
+1· 1−
=
1− 2 .
1−
pi
pi pi
pi
pi
p
p
pi |n
Hence hφ (n) /ni '
i
i
¡P
1
n n2
¢−1
=
6
.
π2
88
4.9
CHAPTER 4. PRIME NUMBERS
Prime numbers and polynomials
We mentioned before that it does not exists a close formula for the value of
prime numbers. However, there is, first of all, an excellent estimate of the
n–th prime, given by
µ
¶
n log log n
pn ' n log n + n (log log n − 1) + O
.
ln n
Q
This expression can be derived by knowing the function (x) which counts
the primes ≤ x.
It is also interesting to notice that there is a remarkable number of polynomial formulas which furnish prime numbers for long sequences of consecutive
integers. Such is, for example, the polynomial
P (x) = x2 + x + 41,
(Euler!) . P (x) is prime for the 40 consecutive values x = 0, 1, 2, ..., 39!
Observe that
Lemma 1. A polynomial with integer coefficients cannot take on prime
values for all integral values of the argument.
Proof. Suppose to the contrary that P (n) is prime for every n. Let
α be an arbitrary positive integer and put β = P (α). Consider now the
sequence
P (α + β) ; P (α + 2β) ; P (α + 3β) ...
By virtue of the binomial theorem, for each n ∈ N
P (α + nβ) − P (α) = integral multiple of β.
Hence P (α + nβ) is also an integral multiple of β and, since this value must
be a prime, it can be only β. Thus
P (α + nβ) − P (α) = P (α + nβ) − β = 0
for every positive integer n , contradicting the fact that a polynomial can
only have finitely many roots.
Other remarkable polynomial expressions are
P (n) = n2 − 79n + 1601,
with P (n) prime for n = 1, 2, ...79.
4.10. DISTRIBUTION OF PRIME NUMBERS
89
P (n) = n2 + n + 17
prime for n = 0, ..., 16,
P (n) = 2n2 + 29,
prime for n = 0, 1, ..., 28.
These quadratic expressions plotted
duce straight lines!
16 15
5 4
6 1
7 8
on a spiral lattice (square spiral) pro14
3
2
9
13
12
.
11
10
What happens on a triangular spiral lattice?
4.10
Distribution of prime numbers
How are prime numbers distribute among the integer ones? This question
was posed for the first time by Gauss and it has given rise to extraordinary
developments in analytic number theory, culminating with the work of Riemann and the prime number theorem by Chebyshev, de la Vallée–Poussin
and Hadamard. The Gauss’s contribution was almost empirical, i.e. he took
the table of primes known at that time and he simply counted the relative
frequency of primes among the integers. It should be noticed that
(i) the primes become rarer and rarer, larger they get
(ii) apart a certain regularity in their mean density, their distribution
seems rather irregular.
In few words, prime numbers resemble the behaviour of statistical mechanics, i.e. smooth properties of the ensemble can be rather irregular for and
particular representative.
The determination of π (x) by Riemann will be the topic of the future lectures. Here we will present a simple argument which permits to derive the
correct asymptotic estimate
π (x) '
x
ln x
It should be noticed that the close formula for the n–th prime number
could be obtained if one knew how to invert the function π (x), i.e.
pn = π −1 (n) .
90
CHAPTER 4. PRIME NUMBERS
FIGURE
The simple proof is based on the observation that the probability that an
arbitrary integer is divisible by the prime pi is 1/pi . Said in another way,
one number over pi is precisely divisible by it (Eratosthenes’s sieve!). Thus,
the probability that it is not divisible by pi is (1 − 1/pi ). Assuming now that
divisibility by different primes is not correlated (we will see that, actually,
it is not the case), the probability that a number x is not divisible by any
prime below it is expressed as
¶
µ
¶µ
¶µ
¶
Yµ
1
1
1
1
W (x) ' 1 −
1−
1−
... '
1−
.
2
3
5
pi
pi ≤x
If x is not divisible by any prime below it, it is of course a prime. Taking
the logarithm of both terms and expanding it for small 1/pi , we have
¶ X
X µ
1
1
ln W (x) '
ln 1 −
'
.
pi
pi
p <x
i
i
The last sum on primes can be expressed as a sum on integers by using the
same function W (x). In fact,
X W (n) Z n W (t)
X 1
'
'
dt.
pi
n
t
1
n
i
Hence, we arrive to the expression
Z
log W (x) ' −
1
x
W (t)
dt.
t
Deriving both terms we obtain the differential equation
1 dW
1
=− ,
W 2 (x) dx
x
whose solution is
W (x) '
1
.
log x
This function expresses the probability that an integer x is a prime. Their
number up to x is then
Z x
Z x
dt
x
Π (x) =
dtW (t) '
≡ Li (x) '
,
ln x
1
1 ln t
4.10. DISTRIBUTION OF PRIME NUMBERS
91
i.e., the result by Gauss.
We will say that the exact expression for the smooth part of Π (x) will be
given by
1 ¡√ ¢ 1 ¡ √ ¢
Π (x) ≡ R (x) = Li (x) − Li x − Li 3 x + ...
2
3
≡
∞
X
µ (n)
n=1
n
³ 1´
Li x n ,
where µ (n) ,n = p1 ...pk is the Moebius coefficient, defined by

n=1
 1
0
...
µ (n) =
.

k
(−1)
otherwise
More surprising is the fact that there exists a close exact formula for Π (x),
able to follow also its jumps. This formula involves the nontrivial zeros of
the Riemann zeta function. Indeed
Π (x) = lim Rk (x) ,
k→∞
where
Rk (x) ≡ R (x) +
k
X
R (xρl ) ,
l=−k
and ρl is the l–th zero of the Riemann zeta function
ζ (s) =
∞
X
1
.
ns
n=1
The first nontrivial zeros have all real part equal to 1/2, whereas their imaginary part is given by
TABLE
92
CHAPTER 4. PRIME NUMBERS
4.11
Probabilistic methods
By using the previous result on the distribution of primes, we can have a
fair estimate of several questions in number theory. In the following we will
discuss some of them.
X 1
→∞
pi
i
since
X W (n) Z Λ dn
X 1
'
'
' log log Λ.
pi
n
1 n ln n
n
i
4.11.1
Coprime probability
What is the probability that two numbers (a, b) randomly chosen are co–
primes?
The probability that one of them is divisible by pi is p1i and the probability that both are divisible by the same prime, assuming no correlations,
is 1/p2i . Hence the probability that they are not both divisible by pi is
¶
Yµ
6
1
P =
1 − 2 = 2.
π
pi
i
Since
¶
¶ X
µ
∞
Yµ
1 −1 Y
1
1
1
π2
1− 2
.
=
1 + 2 + 4 + ... =
=
n2
6
p
p
p
i
i
i
n=1
i
i
4.11.2
Square–free probability
What is the probability that a randomly chosen integer n is square free (i.e.
not divisible by a square)? The probability is equal to the above one, i.e.
P =
6
= 0.608...
π2
In fact, if an integer is square free, it must not be divisible for the same
prime pi more than once. Hence, either it is not divisible by pi or, if it is, it
is not divisible again. Thus
µ
¶
µ
¶
1
1
1
1
Pi ' 1 −
+
1−
= 1 − 2,
pi
pi
pi
pi
and taking the product on all i, we obtain the above result.
4.12. MERSENNE NUMBERS
4.11.3
93
Merten’s formula
The previous estimate of the product on primes can be refined as
¶
Yµ
1
e−γ
1−
'
,
pi
ln x
p <x
i
where γ is the Euler–Mascheroni constant
γ = 0.57721...,
e−γ = 0.5614...
The above number gives, somehow, an estimate of the correlation among
prime numbers. This means that in the previous derivation of the “prime
number theorem”, we should take the product on the primes not less than
x (this would be excessive any way), but
µ
¶
1
1
1−
'
.
pi
ln x
exp(−γ)
Y
pi <x
4.12
Mersenne numbers
These are the primes of the form
Mp = 2p − 1,
where p is a prime (observe that if p were not a prime, say p = n · m, we
would have
³
´
2n·m − 1 = (2n − 1) 2n(m−1) + 2n(m−2) + ...1 ,
i.e. the number is composite). Moreover, not for any prime p is 2p − 1 a
prime number.
Notice that any Mersenne number has as a companion a perfect even number, i.e. a number equal to the sum of its divisors:
P = Mp 2p−1 .
The above formula is a necessary and sufficient condition in order for an
even number to be perfect. It is still an open question whether there exist
odd perfect numbers.
94
CHAPTER 4. PRIME NUMBERS
There exists a remarkable conjecture, due to Wagstaff, concerning these
numbers. Given a number of the form 2p − 1, the probability that it is
prime is about
1
1
'
.
ln (2p − 1)
p ln 2
Among the numbers of this form, the Mersenne’s ones are quite special.
Indeed, it can be shown that if a prime q divides 2p − 1, then q must be of
the form q = kp + 1 for some k ∈ N. So, no “small” primes can divide a
number of the form 2p − 1. Qualitatively, this implies that such numbers
have a bigger probability of being primes than the “prime theorem” estimate.
To quantify it, we will argue on the reverse, namely, the fact that for each
prime q < p, the probability that q divides 2p − 1 is zero. This increases the
³
´−1
probability of 2p − 1 to be a prime by 1 − 1q
. Therefore, for all primes
less than p, we have
¶
Yµ
1 −1
1−
' eγ ln p.
q
q<p
Combining now the two probabilities, together with the one that p is a
prime, we obtain for the probability that 2p − 1 is prime,
eγ
eγ ln p
'
.
ln p ln (2p − 1)
ln 2p
Therefore, for the total number of Mersenne primes less than x we have
Y
eγ X 1
eγ
=
ln ln x.
M ers (x) '
ln 2
p
ln 2
p<ln x
There is a remarkable fit of this formula. This can also be expressed as
follows: If x is the n–th Mersenne number, so that
Y
n=
M ers (x) ,
Then
ln 2
n.
eγ
This conjecture in turn implies, since log log x → ∞, that there exist infinitely many Mersenne primes. But the quantity of Mersenne numbers less
than x os approximately π (log x), i.e. the quantity of primes less than log x
(because log2 x differs only by a constant factor log 2 from log x). Since
log log x '
π (log x) '
log x
log log x
4.13. ARITHMETICAL FUNCTIONS
95
is much bigger than log log x, the Wagstaff conjecture implies also that there
are infinitely many composite Mersenne numbers.
4.13
Arithmetical functions
Sequences of real or complex numbers play a major role in number theory.
They are called arithmetical functions.
Definition 2. An arithmetical function is a function f: N → R or C.
There are many classical examples of arithmetical functions.
Definition 3. An arithmetical function f (n) is said to be multiplicative if
f (n · m) = f (n) · f (m) ,
(n, m) = 1.
The multiplicative function f (n) is said to be completely multiplicative
if
f (n · m) = f (n) · f (m) ,
4.14
f or all n, m ∈ N.
Dirichlet characters
There are also periodic arithmetical functions, which are sequences of real
of complex numbers, periodic with period T :
f (n + T ) = f (n) .
The most important example is provided by the Dirichlet characters.
Definition 4. A Dirichlet character of the conductor T is a function χ (n)
satisfying the following axioms:
1) periodicity: χ (n + T ) = χ (n)
2)multiplicativity: χ (nm) = χ (n) χ (m)
3) χ (n) = 0, if (n, T ) 6= 1.
The principal character χT of conductor T is given by
½
1
if (n, T ) = 1
χT (n) =
.
0
if (n, T ) 6= 1
This definition is motivated by the role played by Dirichet characters in
group theory.
96
CHAPTER 4. PRIME NUMBERS
Chapter 5
The Riemann ζ function
The Riemann ζ (s) function is defined for Res ≥ 1 by the following relation
∞
X
1
.
ns
ζ (s) =
(5.1)
n=1
The series is absolutely convergent for σ > 1, where s = σ+it. The following
result holds:
∞
X
Y
1
1
³
´.
(5.2)
=
s
n
1 − 1s
p
n=1
³
Expanding the factor 1 −
Y
pi
1
psi
pi
i
´−1
in the domain σ > 1, we have
¸
Y·
1
1
1
´=
³
1 + s + 2s + 3s + ... =
pi
pi
pi
1 − p1s
i
1
i
µ
1+
1
1
+ ... + s
ps1
pk
¶
µ
+
¶
1
+ ... + ...
ps1 ps2
Each term of the product is an integer, due to the unique factorization
theorem, hence one obtains eq. (5.2). The above formula (as well as the
others which we will discuss below) is actually a particular example of a
more general formula involving multiplicative functions.
For these functions we have
¡
¢
¡ ¢
f pα1 1 pα2 2 ...pαk k = f (pα1 1 ) f (pα2 2 ) ...f pαk k
97
98
CHAPTER 5. THE RIEMANN ζ FUNCTION
and the general identity
∞
X
f (n)
n=1
ns
=
Y
pi
"
#
¡ ¢
¡ ¢
f p3i
f (pi ) f p2i
+ 2s + 3s + ... .
1+
psi
pi
pi
The identity (5.1) was already established by Euler, but it was Riemann who
had the idea of studying the ζ function in terms of the complex variable s.
It is easy to see that ζ (s) has no zeroes for σ > 1. In fact, for σ > 1, we can
use its product representation of this function and we have
µ
¶µ
¶µ
¶
µ
¶
1
1
1
1
1
1
1− s
1− s
1 − s · ... · 1 − s ζ (s) = 1 + s + s + ...
2
3
4
p
m1 m2
where m1 , m2 ,... are all integers whose prime factors exceed P . Hence
¯µ
¯
¶µ
¶µ
¶
µ
¶
¯
¯
1
1
1
1
1
¯ 1− 1
1− s
1 − s · ... · 1 − s ζ (s)¯¯ ≥→ 1−
>0
σ −...−
¯
s
2
3
4
p
(p + 1)
(p + k)σ
if P is large enough. Therefore |ζ (s)| > 0.
Said in another way, a product can vanish if and only if at least one of the
factors vanishes. Since the general expression is
es ln pi
psi
eσ ln p [cos (t ln p) + i sin (t ln p)]
= s ln p
,
= σ ln p
i − 1
−1
e
e
[cos (t ln p) + i sin (t ln p)] − 1
psi
we see that the generic term never vanishes. On the other hand, the point
s = 1 is singular for ζ (s).
5.1
Relation between ζ (s) and
Q
(x)
There is a deep relation between the behaviour of ζ (s) and the function
π (x), which counts the primes less than x. It can be seen as follows. For
σ > 1 (where Euler’s formula is valid), we have
¶
µ
¶
∞
X µ
X
1
1
log ζ (s) = −
ln 1 − s = −
[π (n) − π (n − 1)] ln 1 − s =
p
n
p
n=2
· µ
¶
µ
¶¸
1
1
−
π (n) ln 1 − s − ln 1 −
=
n
(n + 1)s
n=2
Z n+1
Z ∞
∞
X
s
π (x) dx
π (n)
dx = s
,
s
x (x − 1)
x (xs − 1)
n
2
∞
X
n=2
i.e.
Z
ln ζ (s) = s
2
∞
π (x) dx
.
x (xs − 1)
5.2. MOEBIUS FUNCTION
5.2
99
Moebius function
For σ > 1, we define
¶ X
∞
Yµ
1
1
µ (n)
ZM (s) =
=
1− s =
,
ζ (s)
p
ns
p
(5.3)
n=1
where

n=1
 1
µ (n) =
(−1)k
n = p1 ...pk ,

0
n = p21 ...
pi 6= pj ,
i.e. µ (n) is zero each time that the number n contains more than once the
same factor pi . From this respect, prime numbers appear as “fermionic”
particles!
The arithmetic function µ (n) is known as the Moebius function. It plays a
fundamental role in analytic number theory, since it rules the convolution
of series. Let us study its main properties.
First of all
½
X
1
q=1
µ (d) =
,
0
q>1
d|q
This identity follows from the relations
1 = ZM (s) ζ (s) =
∞
∞
∞
X
1 X µ (n) X 1 X
=
µ (d) .
ms
ns
qs
m=1
n=1
q=1
d|q
The Moebius function enters the so called inversion formulas, i.e.
X
g (q) =
f (d)
d|q
⇓
X ³q´
f (d) =
µ
g (d) .
d
d|q
The above formula can be written in a variety of ways:
∞
∞
³x´
³x´
X
X
f
µ (n) g
g (x) =
↔ f (x) =
,
n
n
n=1
H (x) =
∞
X
n=1
(5.4)
n=1
h (nx) ↔ h (x) =
∞
X
n=1
µ (n) H (nx) .
(5.5)
100
CHAPTER 5. THE RIEMANN ζ FUNCTION
Let us see the proof of (5.4). Making the change of variable x = t/m and
summing on m, we get
∞
X
µ
f
m=1
since
P
5.3
d|q
t
m
¶
=
∞ X
∞
X
µ
µ (n) g
m=1 n=1
t
nm
¶
X µt¶X
=
g
µ (d) = g (x) ,
q
q
d|q
µ (d) = δq,1 .
The Mangoldt function Λ (n)
Taking the logarithm and differentiating eq. (5.2), we have for σ > 1
X ln p
ζ 0 (s)
=−
ζ (s)
ps
p
µ
¶
∞
X
X
1 −1
1
ln p
=
1− s
=−
ms
p
p
p
m=1
=−
∞
X
Λ (n)
n=2
where
½
Λ (n) =
log p
0
ns
,
if n = p or a power of f
.
otherwise
Integrating we have
log ζ (s) =
∞
X
Λ1 (n)
n=2
ns
,
where Λ1 (n) = Λ (n) / ln n.
The Riemann function enters a variety of identities. For instance,
∞
ζ (s − 1) X φ (n)
=
ζ (s)
ns
σ > 2,
n=1
where φ (n) is the Euler function. In fact
ζ (s − 1) Y
=
ζ (s)
p
µ
1 − p−s
1 − p1−s
¶
=
¶µ
¶¾
Y ½µ
1
p
p2
1− s
1 + s + 2s + ...
=
p
p
p
p
µ
¶µ
¶¾
Y½
1
p
p2
1+ 1−
+
+ ...
.
p
ps p2s
p
5.4. ANALYTICITY PROPERTIES OF ζ (S)
101
Observe that, if n = pα1 1 ...pαk k
µ
¶ µ
¶
1
1
φ (n) = n 1 −
... 1 −
.
p1
pk
Hence we obtain the above formula.
We also have
∞
X
X 1
µ (n)
=
log ζ (ns) .
s
p
n
p
n=1
To prove it, notice that
log ζ (s) =
XX
m
with P (s) =
Hence
P
pp
m=1
−s .
∞
X
µ (n)
n=1
p
∞
X
1
P (ms)
=
,
mpms
m
n
∞
∞
X
µ (n) X P (mns)
log ζ (ns) =
=
n
m
n=1
m=1
∞
X
P (rs) X
µ (n) = P (s) .
r
r=2
Exercise. Prove that
n|2
∞ 2
Y
p +1
5
= .
2
p −1
2
p=2
5.4
Analyticity properties of ζ (s) and its functional
equation
The function ζ (s) is regular for all values of s except s = 1, where there is
a simple pole with residue 1. It satisfies the functional equation
ζ (s) = 2s π s−1 sin
πs
Γ (1 − s) ζ (1 − s) .
2
This equation can be written in a more symmetric way: introducing the
function
³s´
s
1
ζ (s)
φ (s) ≡ s (s − 1) π − 2 Γ
2
2
102
CHAPTER 5. THE RIEMANN ζ FUNCTION
we have
φ (s) = φ (1 − s) .
We know that ζ (s) is defined for Res > 1. How can we extend analytically
definition to the complex plane? Consider the Γ function: Γ (s) =
R ∞ itss−1
dxx
e−x . With the substitution x → nx, we have
0
Z ∞
s
Γ (s) = n
dxxs−1 e−nx ,
0
hence
Γ (s)
=
ns
Z
∞
dxxs−1 e−nx .
0
Summing now on n , we have
Z
∞
Γ (s) ζ (s) =
dx
0
xs−1
.
ex − 1
Consider now the integral
Z
I (s) =
C
z s−1
dz,
ez − 1
where the contour C starts at infinity on the positive axis, encircles the
origin once in the positive direction excluding the points ±2πi, ±4πi, . . .
and returns to the positive axis again.
FIGURE
If Res > 1, we can shrink the circle around the origin to zero, so that
Z ∞ s−1
Z ∞ ¡ 2πi ¢s−1
xe
x
I (s) = −
dx +
dx =
x
e −1
ex − 1
0
0
Z
¡ 2πis
¢ ∞ xs−1
¡ 2πis
¢
2πieiπs
e
−1
dx
=
e
−
1
Γ
(s)
ζ
(s)
=
ζ (s) .
ex − 1
Γ (1 − s)
0
Since Γ (s) Γ (1 − s) =
π
sin πs ,
we obtain
e−iπs Γ (1 − s)
ζ (s) =
2πi
Z
C
z s−1
dz.
ez − 1
This formula has been proved for Res > 1. However, the integral is uniformly convergent in any finite region of the s–plane and so it is an integral
5.4. ANALYTICITY PROPERTIES OF ζ (S)
103
function of s. Hence, it defines the analytic continuation of ζ (s) for all
values of s. The only possible singularities are the poles of Γ (1 − s), i.e.
s = 1, 2, 3, .... We know already that ζ (s) is regular at s = 2, 3, 4... and this
follows immediately from Cauchy theorem (the contour, in these cases, does
not encircle any singularities).
The only possible singularity is a simple pole at s = 1. In this case
Z
dz
= 2πi,
I (1) =
z −1
e
C
Γ (1 − s) = −
1
+ ...
s−1
1
,
s−1
s → 1.
Therefore
ζ (s) '
Notice that if s is any integer (negative or null), the integrand in I (s) is
one–valued, and I (s) can be evaluated by the theorem of residues. Since
∞
X
xn
x
=
B
,
n
ex − 1
n!
n=0
we have
1
ζ (0) = − ,
2
ζ (−2m) = 0,
B2m
.
2m
In order to derive the functional equation of ζ (s), consider the integral along
the contour CM
ζ (1 − 2m) = (−1)m
FIGURE
Inside the contour there are now poles at
±2πi, ±4πi, ..., ±2nπi.
The residues at 2mπi and −2mπi are together
³
´
³
´
iπ s−1
3iπ s−1
π (s − 1)
=
2mπe 2
+ 2mπe 2
= (2mπ)s−1 eiπ(s−1) 2 cos
2
−2 (2mπ)s−1 eiπs sin
πs
.
2
104
CHAPTER 5. THE RIEMANN ζ FUNCTION
Hence
Z
I (s) =
CM
n
z s−1
πs X
iπs
dz
+
4πie
sin
(2mπ)s−1 .
ez − 1
2
m=1
Consider now Res < 0, and n
1/ (ex − 1) is bounded on
³ → ∞.´ The function
R
the contour C and z s−1 = 0 |z|s−1 , hence CM → 0, i.e.
I (s) = 4πieiπs sin
∞
πs
πs X
(2mπ)s−1 = 4πieiπs sin
(2π)s−1 ζ (1 − s) ,
2
2
m=1
from which one derives the functional equation.
5.4.1
Consequences of the functional equation
• Link with the Bernoulli numbers:
ζ (2m) = 22m−1 π 2m
B2m
(2m)!
• For the derivative at zero:
1
ζ 0 (0) = − ln 2π.
2
From the functional equation,
−
1
πs Γ0 (s) ζ 0 (s)
ζ 0 (1 − s)
= − ln 2π − π tan
+
+
.
ζ (1 − s)
2
2
Γ (s)
ζ (s)
Γ0 (s)
Γ(s)
1
Near s ' 1, 12 π tan πs
2 ' − s−1 , and
ζ 0 (s)
=
ζ (s)
= −γ, so
1
− (s−1)
2 + A + ...
1
(s−1)
+ γ + A (s − 1)
Hence
−
=−
1
+ γ.
s−1
ζ 0 (0)
= − ln 2π.
ζ (0)
The functional equation for the Riemann function can be derived in many
different ways. One is, for instance, to use the Poisson resummation formula
+∞
X
−∞
f (n) =
+∞ Z
X
−∞
+∞
−∞
f (x) cos 2πnxdx.
5.5. SOME CONSEQUENCES OF EULER’S FORMULA
105
Let us sketch the original proof by Riemann.
¡ ¢
Make the change of variable x → n2 πx in the Γ function Γ 2s :
Z
∞
s−1
2
x
e
−n2 πx
dx =
0
Γ
¡s¢
2
s
ns π 2
.
For Res > 1, summing on n, we have
¡ ¢
Z ∞
Γ 2s ζ (s)
s
=
x 2 −1 ψ (x) dx
s
π2
0
ψ (x) ≡
∞
X
e−n
2 πx
.
n=1
This function is a Jacobi θ–function. It satisfies the basic functional identities
+∞
+∞
X
1 X −n2 π
−n2 πx
x,
e
=√
e
x
−∞
−∞
· µ ¶
¸
1
1
2ψ (x) + 1 = √ 2ψ
+1 .
x
x
After some manipulations, we get
Γ
³s´
2
ζ (s) π
− 2s
Z
1
x
=
s
−1
2
0
1
+
s (s − 1)
Z
Z
ψ (x) dx +
∞
s
x 2 −1 ψ (x) dx =
1
∞h
i
s
s
x− 2 −1 + x 2 −1 ψ (x) dx.
1
The last integral is convergent for all values of s and therefore this formula
remains true, by analytic continuation, for all values of s. But the right
hand side is invariant under the substitution s → 1 − s. Hence
µ
¶
³s´
1−s
− 2s
− 12 + 2s
π Γ
ζ (s) = π
Γ
ζ (1 − s) .
2
2
5.5
Some consequences of Euler’s formula
The Euler’s formula reads
¶
∞ µ
∞
Y
X
1 −1
1
=
1− s
.
ζ (s) =
ns
p
p
n=1
106
CHAPTER 5. THE RIEMANN ζ FUNCTION
The relation with the function π (x):
ln ζ (s) = −
X
p
−
¶
µ
¶
µ
∞
X
1
1
ln 1 − s = −
[π (n) − π (n − 1)] ln 1 − s =
p
n
n=2
∞
X
n=2
=
∞
X
· µ
¶
µ
¶¸
1
1
π (n) ln 1 − s − ln 1 −
=
n
(n + 1)s
Z
n+1
π (n)
n
n=2
Also,
s
dx = s
s
x (x − 1)
Z
∞
2
π (x) dx
.
x (xs − 1)
¶ X
∞
Yµ
µ (n)
1
1
.
=
1− s =
ζ (s)
p
ns
p
n=1
The Moebius function has the property
X
µ (d) = δq,1 .
d|q
It is easy to prove:
1=
∞
∞
∞
X
1 X µ (n) X 1 X
µ (d) .
=
ms
ns
qs
m=1
n=1
q=1
d|q
Moebius inversion formula:
g (x) =
∞
³x´
X
f
n
n=1
f (x) =
∞
X
µ (n) g
n=1
Indeed:
f
³x´
m
=
∞
X
µ (n) g
n=1
³x´
n
.
³ x ´
;
mn
µ ¶
³ x ´ XX
³x´
X ³x´ X
x
f
=
µ (n) g
=
µ (q) g
=g
= g (x) .
m
mn
q
1
m
m.n
q
d|q
Finally, observe that
5.6. DIRICHLET SERIES
107
∞
X
X 1
µ (n)
=
log ζ (ns) ,
s
p
n
p
n=1
with
ln ζ (s) =
XX
m
where F (s) =
Hence,
X µ (n)
n
5.6
P
p
∞
X
F (ms)
1
=
,
ms
mp
m
m=1
1
ps .
ln ζ (ns) =
X µ (n)
n,m
nm
F (mns) =
∞
X
Γ (rs) X
r=1
s
µ (r) = Γ (s) .
n|r
Dirichlet series
Dirichlet series are one of the most relevant and classical topics in analytic
number theory. Here we will follow strictly the book by J.–P. Serre [10].
Let a (n) be a complex–valued arithmetic function.
Definition 5. A Dirichlet series with exponents {λn } is a series of the form
∞
X
a (n) e−λn z ,
(5.6)
n=1
where {λn } is a sequence of real numbers tending to infinity, and z ∈ C.
The simplest example is when λn = n for all n. Then, putting e−z = t, the
series (5.6) is a power series in t.
Another nontrivial case is λn = log n. This reduces the series (5.6) to an
ordinary Dirichlet series.
We remind that, if the coefficients a (n) are bounded, then ϕ (s) is absolutely
convergent for Re(s) > 1.
Definition 6. An (ordinary) Dirichlet series with coefficients a (n) is a
series of the form
∞
X
a (n)
ϕ (s) =
.
ns
n=1
108
CHAPTER 5. THE RIEMANN ζ FUNCTION
The most
known examples of Dirichlet series are the Riemann zeta function
P∞
ζ (s) = n=1 n1s and the L–series
L (s, χ) =
∞
X
χ (n)
n=1
ns
,
corresponding to a Dirichlet character χ of conductor T . Other interesting
example are
∞
ζ (s − 1) X φ (n)
,
=
ζ (s)
ns
n=1
where φ (n) is the totient function, and
ζ (s) ζ (s − a) =
∞
X
σa (n)
n=1
ns
,
where σa (n) is the divisor function. Also, the logarithm of the zeta function
is expressed, for Re(s) > 1, by the Dirichlet series
log ζ (s) =
∞
X
Λ (n) 1
,
log (n) ns
n=2
where Λ (n) is the von Mangoldt function. A very simple result holds for the
P
f (n)
logarithmic derivative of a Dirichlet series Φ (s) = ∞
n=1 ns , corresponding
to a completely multiplicative function f (n):
∞
X f (n) Λ (n)
Φ0 (s)
=−
.
Φ (s)
ns
n=1
Chapter 6
Numerical Spectral Methods
Most of the times, the Schrödinger equation cannot be solved exactly. In
order to determine the energy spectrum, one has to rely on some numerical
methods. Here we will discuss, in particular, two different approaches. Each
of them has certain advantages and disadvantages.
The first of them is the truncated Hilbert space method, the second one the
tight–binding method. Before starting the discussion, it is useful to remind
some basic formulas concerning the harmonic oscillator.
6.1
Basic formulas for the quantum harmonic oscillator
The Hamiltonian is
1
p2
+ mω 2 q 2 =
2m 2
³
2
mω
p ´³
p ´ ~ω
q+i
q−i
−
,
2
mω
mω
2
H=
with
[q, p] = i~.
Introducing
r
a=
we get
r
mω ³
p ´
q+i
,
2~
mω
+
a =
£
¤
a, a+ = 1.
109
mω ³
p ´
q−i
2~
mω
110
Hence
CHAPTER 6. NUMERICAL SPECTRAL METHODS
r
¢
¢
~ ¡
l ¡
q=
a + a+ = √ a + a+ ,
2mω
2
r
¢ i~ 1 ¡
¢
mω~ ¡ +
p=i
a − a = √ a+ − a ,
2
l 2
where l is the basic length
r
~
l=
.
mω
In terms of a and a+ , the Hamiltonian can be written as
µ
¶
1
+
H = ~ω a a +
.
2
Denoting by |ni the n–th eigenvector of this Hamiltonian, we have
√
a |ni = n |n − 1i ,
√
a+ |ni = n + 1 |n + 1i ,
The Hilbert space of physical states is finitely generated, since it can be
obtained from the vacuum state via the formula
1 ¡ + ¢n
a
|ni = √
|0i .
n!
The occupation number operator is defined by
N = a+ a,
and for its eigenstates we have
N |ni = n |ni .
Finally, we recall that the Schrödinger representation is defined by
µ
¶
~ d
aψ0 (q) = q +
ψ0 (q) = 0,
mω dq
i.e.
µ
¶
³ mω ´ 1
mωq 2
4
exp −
.
ψ0 (q) =
~π
2~
The excited states are expressed by
r
µr
¶
1 ³ mω ´ 41 − mω q2
mω
e 2~ Hn
ψn (q) =
q ,
2n n! ~π
~
where the operator
dn −x2
e
dxn
is the generator of the classical Hermite polynomials.
2
Hn (x) = (−1)n ex
(6.1)
6.2. TRUNCATED HILBERT SPACE METHOD
6.2
111
Truncated Hilbert space method
Let us assume for simplicity that the potential V (x) entering the Schrödinger
equation is an even function, unbounded at infinity
FIGURE
The problem is to find the eigenvalues and the eigenvectors of the Hamiltonian
p2
H=
+ V (x) .
2m
The THSM, first of all, amounts to choosing a basis |ψn i in the Hilbert space
and computing the matrix elements
hψn | H |ψm i ≡ Hn,m .
Second, one has to truncate the above basis at a certain level N and to
consider the finite–dimensional Hamiltonian
HN = Hn,m
n, m = 0, 1, ..., N
which can be diagonalized numerically. A practical way of implementing the
algorithm consists of using the harmonic oscillator basis (6.1). Its vectors
depend on the free parameter ω. The strategy then is the following.
(i) Use the variational principle on the ground state wave function to
identify the optimal value of ω, say ω, and use this value to construct the
remaining vectors of the basis.
p2
(ii) Compute separately the matrix elements of the kinetic term T = 2m
and of the potential term V (x).
(iii) On the harmonic oscillator basis, T has matrix elements on the
diagonal, and along the next main diagonals only

..
.
..

.

 ... ...


..

.

T =




0

..
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
..
0
..









.. 
. 

..
.
112
CHAPTER 6. NUMERICAL SPECTRAL METHODS
as evident from the representation p.
(iv) The potential V (x), in general, has sparse matrix elements, given
by all the overlapping integrals
Z
Vnm = dxψn (x) V (x) ψm .
The main advantages of the method are:
1) It is easy to implement.
2) It furnishes reasonable estimates of the first eigenvalues.
The main disadvantages are:
1) The integrals Vnm can be only poorly estimated for large values of the
quantum numbers n and m, due to the oscillatory nature of the Hermite
polynomials (the same is true for any other basis).
2) The final form of the truncated Hamiltonian HN , as mentioned above,
in general possesses nonzero matrix elements in all entries. For large values of
N this may represent a problem for a fast and sufficiently accurate numerical
diagonalization of this matrix.
However, for polynomial potentials, one can overcome at least the first of
the mentioned disadvantages, since the algebraic expressions of the matrix
elements can be computed without performing the numerical integration.
Let’s see how this is possible.
Consider the Hamiltonian
p2
+ λ |x|n ,
2m
where n is an integer. First of all, we can identify the optimal choice of ω for
this Hamiltonian and, correspondingly, the harmonic oscillator length
r
~
l=
.
mω
H=
The related representation of the operators x and p reads
¢
l ¡
x = √ a + a+ ,
2
¢
i~ 1 ¡
p = √ a+ − a .
l 2
On the other hand, the above Hamiltonian can be put into and a–dimensionless
form by using the intrinsic length scale of the problem
1
µ 2
¶ n+2
~
ξ=
b (n)
,
mλ
6.2. TRUNCATED HILBERT SPACE METHOD
113
where the function b (n), as before, is chosen so that the energy levels match,
for large values of the quantum numbers, with their semi–classical expressions.
We are led to compute the universal ration of the two scales
l
R= ,
ξ
and by substituting the value of the optimal choice of ω,
Ã
ωop =
we obtain
Ã
R=
2nΓ
¡ n+1 ¢
λ~
√2 n
πm 2
n−2
2
2
! n+2
,
! 1
√
n+2
π
¡ n+1 ¢
.
2nb (n) Γ 2
The Hamiltonian can be written as
·
¸
¢2 bRn ¡
¢
1 1 ¡ +
+ n
H = ξ0 −
a −a + n a+a
.
4 R2
22
The operators a and a+ can

0
 0

 0

a= 0


 0





a+ = 



(6.2)
be easily computed, since
√

1 0√ 0
···

0
2 0√


0
0
3 √
0


0
0
0
4

.. 
. 
0
0
0
0
0√
1
0
0
0
0
0√
2
0
0
0
0
0√
3
0
0
0
0
0
√
4

···




.



0
0
0
..
.
Obviously, the truncated matrices a(N ) and a+
(N ) do not satisfy the usual
commutation relation
£
¤
a, a+ = I.
114
CHAPTER 6. NUMERICAL SPECTRAL METHODS
Instead, they satisfy the relation
h
i
a(N ) , a+
(N ) = IN − (N + 1) δN,N ,
i.e.



h
i 

a(N ) , a+
(N ) = 




1




,



1
1
1
..
.
−N
so that the trace of the commutator vanishes, as it happens for any finite
dimensional representation. Even with this drawback, it is however evident
the algebraic advantage of the expression (6.2). In fact, the matrix elements
of this matrix can be computed just by sum and matrix multiplication of
the known matrices aN and a+
N.
It still remains the computational problem of diagonalizing a sparse matrix.
This problem can be attacked by using a different method, the so–called
Tight–Binding Method.
6.3
Tight–Binding Method
This method reduces the problem of finding the eigenvalues and the eigenvectors of the Schrödinger Hamiltonian to the one of diagonalizing numerically
a matrix with non–vanishing matrix elements only along the main diagonal and the upper and lower ones. However, a certain care is needed in
order to interpret correctly the results. To this aim, it is useful to discuss
preliminarily the problem of a particle in a box.
6.3.1
Particle in a box
Consider the Schrödinger problem for a free particle in a box of length L:
−
~2 d2
ψ = Eψ,
2m dx2
ψ (0) = ψ (L) = 0.
FIGURE
6.3. TIGHT–BINDING METHOD
115
The solution of the problem is well known. We have
d2 ψ
2m
= − 2 Eψ,
2
dx
~
with
Ãr
ψ = A sin
!
2mE
x+ϕ .
~2
³q
By imposing the boundary conditions, we end up with ϕ = 0 and sin
0, i.e.
r
2mE
L = nπ
(n = ±1, ±2, ...)
~2
~2 π 2 2
n ,
n = 1, 2, ...
2mL2
The (normalized) wave functions are given by
r
³ nπ ´
2
sin
x .
ψn =
L
L
En =
2mE
L
~2
(6.3)
Notice that ψn and ψ−n describe the same physical vector. Suppose now
that we discretize the previous equation, i.e. first we divide the interval
(0, L) in N steps, as in figure
FIGURE
with lattice space
L
.
N
The wave functions take now values only on the discrete points pn ≡ nε,
with ψ0 = ψN = 0. The discrete version of the second derivative term is
ε≡
ψm+1 + ψm−1 − 2ψm
d2 ψ (x)
→
.
dx2
ε2
In this way we are led to find the eigenvalues and eigenvectors of the following
linear problem
µ
¶
2mε2
ψm+1 + ψm−1 − 2ψm = − 2
Eψm .
(6.4)
~
This problem can be easily solved by expressing ψm as
´
=
116
CHAPTER 6. NUMERICAL SPECTRAL METHODS
ψm =
X
eikmε ψbk ,
k
where the range of k will be clear from the subsequent considerations.
Substituting this expression into (6.4), we have
2
2mε
2 cos (kε − 1) ψbk = − 2 E ψbk ,
~
i.e.
~2 1
(1 − cos kε) .
(6.5)
m ε2
For an infinitesimal value of ε we shall recover the previous result: expanding
the right hand side of (6.5) we have
E (k) =
E (k) '
~2 k 2
,
m 2
(6.6)
and comparing with (6.3) we obtain the discrete values of k:
kn =
πn
,
L
n = 0, ±1, ...
Since L = nε, the only relevant values are
kn =
πn
,
nε
n = 0, ±1, ... ± (N − 1) .
However, due to the parity of the expression (6.5) for k → −k, and taking
into account the boundary conditions
ψ0 = ψN = 0,
we can restrict the attention to the values
πn
kn =
,
n = 0, 1, ..., N − 1,
L
so that, finally we get
En =
~2 N 2 ³
πn ´
,
1
−
cos
m L2
N
n = 1, 2, ..., N − 1.
For the corresponding wave functions we have
(n)
ψmε
=
n−1
X
p=0
sin
h πpm i
N
ψbp(n) ,
6.3. TIGHT–BINDING METHOD
where
117
r
2
δn,p .
Nε
The important thing to notice is that now the spectrum of the discrete
Schrödinger equation has additional features, namely the lowest part of the
spectrum is similar to the actual Schrödinger equation, whereas the higher
part of the spectrum is dominated by the lattice effects and it shows the
typical form of the case
ψbp(n) =
FIGURE
Consider now the discrete version of the Schrödinger equation in the
presence of a potential
−
~2 1
(ψm+1 + ψm−1 − 2ψm ) + Vm ψm = Eψm .
2m ε2
For studying the problem, we have to cut the potential on a finite interval
(0, L). This introduces new aspects in the problem: with boundary conditions
ψ0 = ψL = 0.
The effective potential felt by the particle is V (x) inside the box, but V =
∞ at x = 0 and x = L.
FIGURE
It is obvious that if we discretize the Schrödinger equation in terms of N
points, we will obtain N different energy levels.
However, not all of them are those relative to the potential V ! Only those
which take values
E > V0 = VL
may have some resemblance to the actual eigenvalues. The higher ones, in
fact, come just from the infinite well problem, further masked by the lattice
effects. So, typically the spectrum obtained in this way has three different
behaviours, as shown in figure
FIGURE
The region I, the lowest part of the spectrum, probes the potential and
gives an approximation of the true eigenvalues. The region II, where the
118
CHAPTER 6. NUMERICAL SPECTRAL METHODS
eigenvalues are larger that the value of the potential at the edge of the
interval, is simply the spectrum of the infinite well potential, for values of
the discrete index n which allows the expansion (6.6). The region III is,
instead, just a lattice artifact of the spectrum.
In order to separate the above region one has to perform a finite size
scaling, i.e. follow the evolution of the eigenvalues by changing the value of
L (increasing correspondingly the number N of discrete points), the “true”
eigenvalues are those which do not change by varying L.
FIGURE
In conclusion, the tight binding method provides an Hamiltonian which
is relatively simple and fast to diagonalize numerically, given its (almost)
diagonal form


.. ..
.
.


 ... ... ...

0




.. .. ..


.
.
.




.. .. ..


.
.
.



.. .. .. 

.
.
. 
0


.. ..
.
.
The potential is treated in a local way and it is not necessary to compute
the overlapping integrals. However, it is necessary to have certain care
in interpreting the result and in identifying the ”true” eigenvalues of the
problem.
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119